mod/mod_kth_root.hpp

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• Last update: 2024-10-18 02:58:53+09:00
• Include: #include "mod/mod_kth_root.hpp"

Depends on

• ds/hashmap.hpp
• mod/barrett.hpp
• mod/mod_inv.hpp
• mod/mod_pow.hpp
• mod/mongomery_modint.hpp
• mod/primitive_root.hpp
• nt/factor.hpp
• nt/primetest.hpp
• random/base.hpp

Verified with

• test/2_library_checker/number_theory/kth_root_mod.test.cpp

Code

#include "nt/primetest.hpp"

#include "mod/primitive_root.hpp"

#include "mod/mod_inv.hpp"

#include "ds/hashmap.hpp"


// mod は int

int mod_kth_root(ll k, ll a, int mod) {
  assert(primetest(mod) && 0 <= a && a < mod);
  if (k == 0) return (a == 1 ? 1 : -1);
  if (a == 0) return 0;
  if (mod == 2) return a;
  k %= mod - 1;
  Barrett bt(mod);

  ll g = gcd(k, mod - 1);
  if (mod_pow(a, (mod - 1) / g, mod) != 1) return -1;

  ll c = mod_inv(k / g, (mod - 1) / g);
  a = mod_pow(a, c, mod);
  k = (k * c) % (mod - 1);
  if (k == 0) return 1;

  g = primitive_root(mod);

  auto solve_pp = [&](ll p, int e, ll a) -> ll {
    int f = 0;
    ll pf = 1;
    while ((mod - 1) % (pf * p) == 0) ++f, pf *= p;
    ll m = (mod - 1) / pf;
    /*
    ・位数 Qm の巡回群
    ・a の p^e 乗根をとりたい。持つことは分かっている
    ・a / x^{p^e} = b を維持する。まずは、b が p で割れる回数を増やしていく。
    */
    ll x = 1, b = a, c = f - e; // b ^ {mp^c} = 1

    int pc = 1;
    FOR(c) pc *= p;
    int pe = 1;
    FOR(e) pe *= p;
    // 必要ならば原始 p 乗根に関する離散対数問題のセットアップ

    ll G = mod_pow(g, (mod - 1) / p, mod);
    int M = 0;
    HashMap<int> MP;
    ll GM_inv = -1;
    if (c) {
      while (M * M < p) ++M;
      MP.build(M);
      ll Gpow = 1;
      FOR(m, M) {
        MP[Gpow] = m;
        Gpow = bt.mul(Gpow, G);
      }
      GM_inv = mod_pow(Gpow, mod - 2, mod);
    }

    while (c) {
      /*
      b^{mp^c} = 1 が分かっている。(b/x^{p^e}})^{mp^{c-1}} = 1 にしたい。
      x = g^{p^{f-c-e}*k} として探す。原始 p 乗根 B, G に対する B = G^k に帰着。
      */
      ll B = mod_pow(b, m * pc / p, mod);
      int k = [&](ll B) -> int {
        FOR(m, M + 1) {
          if (MP.count(B)) return m * M + MP[B];
          B = bt.mul(B, GM_inv);
        }
        return -1;
      }(B);
      x = bt.mul(x, mod_pow(g, pf / pc / pe * k, mod));
      ll exp = pf / pc * k % (mod - 1);
      b = bt.mul(b, mod_pow(g, mod - 1 - exp, mod));
      --c;
      pc /= p;
    }
    int k = pe - mod_inv(m, pe);
    k = (k * m + 1) / pe;
    ll y = mod_pow(b, k, mod);
    x = bt.mul(x, y);
    return x;
  };

  auto pf = factor(k);
  for (auto&& [p, e]: pf) a = solve_pp(p, e, a);
  return a;
}

ll mod_kth_root_64(ll k, ll a, ll mod) {
  assert(primetest(mod) && 0 <= a && a < mod);
  if (k == 0) return (a == 1 ? 1 : -1);
  if (a == 0) return 0;
  if (mod == 2) return a;
  k %= mod - 1;

  ll g = gcd(k, mod - 1);
  if (mod_pow_64(a, (mod - 1) / g, mod) != 1) return -1;

  ll c = mod_inv(k / g, (mod - 1) / g);
  a = mod_pow_64(a, c, mod);
  k = i128(k) * c % (mod - 1);
  if (k == 0) return 1;

  g = primitive_root_64(mod);

  auto solve_pp = [&](ll p, ll e, ll a) -> ll {
    ll f = 0;
    ll pf = 1;
    while (((mod - 1) / pf) % p == 0) ++f, pf *= p;
    ll m = (mod - 1) / pf;
    /*
    ・位数 Qm の巡回群
    ・a の p^e 乗根をとりたい。持つことは分かっている
    ・a / x^{p^e} = b を維持する。まずは、b が p で割れる回数を増やしていく。
    */
    ll x = 1, b = a, c = f - e; // b ^ {mp^c} = 1

    ll pc = 1;
    FOR(c) pc *= p;
    ll pe = 1;
    FOR(e) pe *= p;
    // 必要ならば原始 p 乗根に関する離散対数問題のセットアップ

    ll G = mod_pow_64(g, (mod - 1) / p, mod);
    ll M = 0;
    ll GM_inv = -1;
    HashMap<ll> MP;
    if (c) {
      while (M * M < p) ++M;
      MP.build(M);
      ll Gpow = 1;
      FOR(m, M) {
        MP[Gpow] = m;
        Gpow = i128(Gpow) * G % mod;
      }
      GM_inv = mod_pow_64(Gpow, mod - 2, mod);
    }

    while (c) {
      /*
      b^{mp^c} = 1 が分かっている。(b/x^{p^e}})^{mp^{c-1}} = 1 にしたい。
      x = g^{p^{f-c-e}*k} として探す。原始 p 乗根 B, G に対する B = G^k に帰着。
      */
      ll B = mod_pow_64(b, pc / p * m, mod);
      ll k = [&](ll B) -> ll {
        FOR(m, M + 1) {
          if (MP.count(B)) return m * M + MP[B];
          B = i128(B) * GM_inv % mod;
        }
        return -1;
      }(B);
      x = i128(x) * mod_pow_64(g, pf / pc / pe * k, mod) % mod;

      ll exp = pf / pc * i128(k) % (mod - 1);
      b = i128(b) * mod_pow_64(g, mod - 1 - exp, mod) % mod;
      --c;
      pc /= p;
    }
    ll k = pe - mod_inv(m, pe);
    k = (i128(k) * m + 1) / pe;
    ll y = mod_pow_64(b, k, mod);
    x = i128(x) * y % mod;
    return x;
  };

  auto pf = factor(k);
  for (auto&& [p, e]: pf) a = solve_pp(p, e, a);
  return a;
}

#line 2 "mod/mongomery_modint.hpp"

// odd mod.
// x の代わりに rx を持つ
template <int id, typename U1, typename U2>
struct Mongomery_modint {
  using mint = Mongomery_modint;
  inline static U1 m, r, n2;
  static constexpr int W = numeric_limits<U1>::digits;

  static void set_mod(U1 mod) {
    assert(mod & 1 && mod <= U1(1) << (W - 2));
    m = mod, n2 = -U2(m) % m, r = m;
    FOR(5) r *= 2 - m * r;
    r = -r;
    assert(r * m == U1(-1));
  }
  static U1 reduce(U2 b) { return (b + U2(U1(b) * r) * m) >> W; }

  U1 x;
  Mongomery_modint() : x(0) {}
  Mongomery_modint(U1 x) : x(reduce(U2(x) * n2)){};
  U1 val() const {
    U1 y = reduce(x);
    return y >= m ? y - m : y;
  }
  mint &operator+=(mint y) {
    x = ((x += y.x) >= m ? x - m : x);
    return *this;
  }
  mint &operator-=(mint y) {
    x -= (x >= y.x ? y.x : y.x - m);
    return *this;
  }
  mint &operator*=(mint y) {
    x = reduce(U2(x) * y.x);
    return *this;
  }
  mint operator+(mint y) const { return mint(*this) += y; }
  mint operator-(mint y) const { return mint(*this) -= y; }
  mint operator*(mint y) const { return mint(*this) *= y; }
  bool operator==(mint y) const {
    return (x >= m ? x - m : x) == (y.x >= m ? y.x - m : y.x);
  }
  bool operator!=(mint y) const { return not operator==(y); }
  mint pow(ll n) const {
    assert(n >= 0);
    mint y = 1, z = *this;
    for (; n; n >>= 1, z *= z)
      if (n & 1) y *= z;
    return y;
  }
};

template <int id>
using Mongomery_modint_32 = Mongomery_modint<id, u32, u64>;
template <int id>
using Mongomery_modint_64 = Mongomery_modint<id, u64, u128>;
#line 3 "nt/primetest.hpp"

bool primetest(const u64 x) {
  assert(x < u64(1) << 62);
  if (x == 2 or x == 3 or x == 5 or x == 7) return true;
  if (x % 2 == 0 or x % 3 == 0 or x % 5 == 0 or x % 7 == 0) return false;
  if (x < 121) return x > 1;
  const u64 d = (x - 1) >> lowbit(x - 1);

  using mint = Mongomery_modint_64<202311020>;

  mint::set_mod(x);
  const mint one(u64(1)), minus_one(x - 1);
  auto ok = [&](u64 a) -> bool {
    auto y = mint(a).pow(d);
    u64 t = d;
    while (y != one && y != minus_one && t != x - 1) y *= y, t <<= 1;
    if (y != minus_one && t % 2 == 0) return false;
    return true;
  };
  if (x < (u64(1) << 32)) {
    for (u64 a: {2, 7, 61})
      if (!ok(a)) return false;
  } else {
    for (u64 a: {2, 325, 9375, 28178, 450775, 9780504, 1795265022}) {
      if (!ok(a)) return false;
    }
  }
  return true;
}
#line 2 "mod/primitive_root.hpp"

#line 2 "nt/factor.hpp"

#line 2 "random/base.hpp"

u64 RNG_64() {
  static u64 x_ = u64(chrono::duration_cast<chrono::nanoseconds>(chrono::high_resolution_clock::now().time_since_epoch()).count()) * 10150724397891781847ULL;
  x_ ^= x_ << 7;
  return x_ ^= x_ >> 9;
}

u64 RNG(u64 lim) { return RNG_64() % lim; }

ll RNG(ll l, ll r) { return l + RNG_64() % (r - l); }
#line 5 "nt/factor.hpp"

template <typename mint>
ll rho(ll n, ll c) {
  assert(n > 1);
  const mint cc(c);
  auto f = [&](mint x) { return x * x + cc; };
  mint x = 1, y = 2, z = 1, q = 1;
  ll g = 1;
  const ll m = 1LL << (__lg(n) / 5);
  for (ll r = 1; g == 1; r <<= 1) {
    x = y;
    FOR(r) y = f(y);
    for (ll k = 0; k < r && g == 1; k += m) {
      z = y;
      FOR(min(m, r - k)) y = f(y), q *= x - y;
      g = gcd(q.val(), n);
    }
  }
  if (g == n) do {
      z = f(z);
      g = gcd((x - z).val(), n);
    } while (g == 1);
  return g;
}

ll find_prime_factor(ll n) {
  assert(n > 1);
  if (primetest(n)) return n;
  FOR(100) {
    ll m = 0;
    if (n < (1 << 30)) {
      using mint = Mongomery_modint_32<20231025>;
      mint::set_mod(n);
      m = rho<mint>(n, RNG(0, n));
    } else {
      using mint = Mongomery_modint_64<20231025>;
      mint::set_mod(n);
      m = rho<mint>(n, RNG(0, n));
    }
    if (primetest(m)) return m;
    n = m;
  }
  assert(0);
  return -1;
}

// ソートしてくれる
vc<pair<ll, int>> factor(ll n) {
  assert(n >= 1);
  vc<pair<ll, int>> pf;
  FOR(p, 2, 100) {
    if (p * p > n) break;
    if (n % p == 0) {
      ll e = 0;
      do { n /= p, e += 1; } while (n % p == 0);
      pf.eb(p, e);
    }
  }
  while (n > 1) {
    ll p = find_prime_factor(n);
    ll e = 0;
    do { n /= p, e += 1; } while (n % p == 0);
    pf.eb(p, e);
  }
  sort(all(pf));
  return pf;
}

vc<pair<ll, int>> factor_by_lpf(ll n, vc<int>& lpf) {
  vc<pair<ll, int>> res;
  while (n > 1) {
    int p = lpf[n];
    int e = 0;
    while (n % p == 0) {
      n /= p;
      ++e;
    }
    res.eb(p, e);
  }
  return res;
}
#line 2 "mod/mod_pow.hpp"

#line 2 "mod/barrett.hpp"

// https://github.com/atcoder/ac-library/blob/master/atcoder/internal_math.hpp
struct Barrett {
  u32 m;
  u64 im;
  explicit Barrett(u32 m = 1) : m(m), im(u64(-1) / m + 1) {}
  u32 umod() const { return m; }
  u32 modulo(u64 z) {
    if (m == 1) return 0;
    u64 x = (u64)(((unsigned __int128)(z)*im) >> 64);
    u64 y = x * m;
    return (z - y + (z < y ? m : 0));
  }
  u64 floor(u64 z) {
    if (m == 1) return z;
    u64 x = (u64)(((unsigned __int128)(z)*im) >> 64);
    u64 y = x * m;
    return (z < y ? x - 1 : x);
  }
  pair<u64, u32> divmod(u64 z) {
    if (m == 1) return {z, 0};
    u64 x = (u64)(((unsigned __int128)(z)*im) >> 64);
    u64 y = x * m;
    if (z < y) return {x - 1, z - y + m};
    return {x, z - y};
  }
  u32 mul(u32 a, u32 b) { return modulo(u64(a) * b); }
};

struct Barrett_64 {
  u128 mod, mh, ml;

  explicit Barrett_64(u64 mod = 1) : mod(mod) {
    u128 m = u128(-1) / mod;
    if (m * mod + mod == u128(0)) ++m;
    mh = m >> 64;
    ml = m & u64(-1);
  }

  u64 umod() const { return mod; }

  u64 modulo(u128 x) {
    u128 z = (x & u64(-1)) * ml;
    z = (x & u64(-1)) * mh + (x >> 64) * ml + (z >> 64);
    z = (x >> 64) * mh + (z >> 64);
    x -= z * mod;
    return x < mod ? x : x - mod;
  }

  u64 mul(u64 a, u64 b) { return modulo(u128(a) * b); }
};
#line 5 "mod/mod_pow.hpp"

u32 mod_pow(int a, ll n, int mod) {
  assert(n >= 0);
  a = ((a %= mod) < 0 ? a + mod : a);
  if ((mod & 1) && (mod < (1 << 30))) {
    using mint = Mongomery_modint_32<202311021>;
    mint::set_mod(mod);
    return mint(a).pow(n).val();
  }
  Barrett bt(mod);
  int r = 1;
  while (n) {
    if (n & 1) r = bt.mul(r, a);
    a = bt.mul(a, a), n >>= 1;
  }
  return r;
}

u64 mod_pow_64(ll a, ll n, u64 mod) {
  assert(n >= 0);
  a = ((a %= mod) < 0 ? a + mod : a);
  if ((mod & 1) && (mod < (u64(1) << 62))) {
    using mint = Mongomery_modint_64<202311021>;
    mint::set_mod(mod);
    return mint(a).pow(n).val();
  }
  Barrett_64 bt(mod);
  ll r = 1;
  while (n) {
    if (n & 1) r = bt.mul(r, a);
    a = bt.mul(a, a), n >>= 1;
  }
  return r;
}
#line 6 "mod/primitive_root.hpp"

// int

int primitive_root(int p) {
  auto pf = factor(p - 1);
  auto is_ok = [&](int g) -> bool {
    for (auto&& [q, e]: pf)
      if (mod_pow(g, (p - 1) / q, p) == 1) return false;
    return true;
  };
  while (1) {
    int x = RNG(1, p);
    if (is_ok(x)) return x;
  }
  return -1;
}

ll primitive_root_64(ll p) {
  auto pf = factor(p - 1);
  auto is_ok = [&](ll g) -> bool {
    for (auto&& [q, e]: pf)
      if (mod_pow_64(g, (p - 1) / q, p) == 1) return false;
    return true;
  };
  while (1) {
    ll x = RNG(1, p);
    if (is_ok(x)) return x;
  }
  return -1;
}
#line 2 "mod/mod_inv.hpp"

// long でも大丈夫

// (val * x - 1) が mod の倍数になるようにする

// 特に mod=0 なら x=0 が満たす

ll mod_inv(ll val, ll mod) {
  if (mod == 0) return 0;
  mod = abs(mod);
  val %= mod;
  if (val < 0) val += mod;
  ll a = val, b = mod, u = 1, v = 0, t;
  while (b > 0) {
    t = a / b;
    swap(a -= t * b, b), swap(u -= t * v, v);
  }
  if (u < 0) u += mod;
  return u;
}
#line 2 "ds/hashmap.hpp"

// u64 -> Val

template <typename Val>
struct HashMap {
  // n は入れたいものの個数で ok

  HashMap(u32 n = 0) { build(n); }
  void build(u32 n) {
    u32 k = 8;
    while (k < n * 2) k *= 2;
    cap = k / 2, mask = k - 1;
    key.resize(k), val.resize(k), used.assign(k, 0);
  }

  // size を保ったまま. size=0 にするときは build すること.

  void clear() {
    used.assign(len(used), 0);
    cap = (mask + 1) / 2;
  }
  int size() { return len(used) / 2 - cap; }

  int index(const u64& k) {
    int i = 0;
    for (i = hash(k); used[i] && key[i] != k; i = (i + 1) & mask) {}
    return i;
  }

  Val& operator[](const u64& k) {
    if (cap == 0) extend();
    int i = index(k);
    if (!used[i]) { used[i] = 1, key[i] = k, val[i] = Val{}, --cap; }
    return val[i];
  }

  Val get(const u64& k, Val default_value) {
    int i = index(k);
    return (used[i] ? val[i] : default_value);
  }

  bool count(const u64& k) {
    int i = index(k);
    return used[i] && key[i] == k;
  }

  // f(key, val)

  template <typename F>
  void enumerate_all(F f) {
    FOR(i, len(used)) if (used[i]) f(key[i], val[i]);
  }

private:
  u32 cap, mask;
  vc<u64> key;
  vc<Val> val;
  vc<bool> used;

  u64 hash(u64 x) {
    static const u64 FIXED_RANDOM = std::chrono::steady_clock::now().time_since_epoch().count();
    x += FIXED_RANDOM;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return (x ^ (x >> 31)) & mask;
  }

  void extend() {
    vc<pair<u64, Val>> dat;
    dat.reserve(len(used) / 2 - cap);
    FOR(i, len(used)) {
      if (used[i]) dat.eb(key[i], val[i]);
    }
    build(2 * len(dat));
    for (auto& [a, b]: dat) (*this)[a] = b;
  }
};
#line 5 "mod/mod_kth_root.hpp"

// mod は int

int mod_kth_root(ll k, ll a, int mod) {
  assert(primetest(mod) && 0 <= a && a < mod);
  if (k == 0) return (a == 1 ? 1 : -1);
  if (a == 0) return 0;
  if (mod == 2) return a;
  k %= mod - 1;
  Barrett bt(mod);

  ll g = gcd(k, mod - 1);
  if (mod_pow(a, (mod - 1) / g, mod) != 1) return -1;

  ll c = mod_inv(k / g, (mod - 1) / g);
  a = mod_pow(a, c, mod);
  k = (k * c) % (mod - 1);
  if (k == 0) return 1;

  g = primitive_root(mod);

  auto solve_pp = [&](ll p, int e, ll a) -> ll {
    int f = 0;
    ll pf = 1;
    while ((mod - 1) % (pf * p) == 0) ++f, pf *= p;
    ll m = (mod - 1) / pf;
    /*
    ・位数 Qm の巡回群
    ・a の p^e 乗根をとりたい。持つことは分かっている
    ・a / x^{p^e} = b を維持する。まずは、b が p で割れる回数を増やしていく。
    */
    ll x = 1, b = a, c = f - e; // b ^ {mp^c} = 1

    int pc = 1;
    FOR(c) pc *= p;
    int pe = 1;
    FOR(e) pe *= p;
    // 必要ならば原始 p 乗根に関する離散対数問題のセットアップ

    ll G = mod_pow(g, (mod - 1) / p, mod);
    int M = 0;
    HashMap<int> MP;
    ll GM_inv = -1;
    if (c) {
      while (M * M < p) ++M;
      MP.build(M);
      ll Gpow = 1;
      FOR(m, M) {
        MP[Gpow] = m;
        Gpow = bt.mul(Gpow, G);
      }
      GM_inv = mod_pow(Gpow, mod - 2, mod);
    }

    while (c) {
      /*
      b^{mp^c} = 1 が分かっている。(b/x^{p^e}})^{mp^{c-1}} = 1 にしたい。
      x = g^{p^{f-c-e}*k} として探す。原始 p 乗根 B, G に対する B = G^k に帰着。
      */
      ll B = mod_pow(b, m * pc / p, mod);
      int k = [&](ll B) -> int {
        FOR(m, M + 1) {
          if (MP.count(B)) return m * M + MP[B];
          B = bt.mul(B, GM_inv);
        }
        return -1;
      }(B);
      x = bt.mul(x, mod_pow(g, pf / pc / pe * k, mod));
      ll exp = pf / pc * k % (mod - 1);
      b = bt.mul(b, mod_pow(g, mod - 1 - exp, mod));
      --c;
      pc /= p;
    }
    int k = pe - mod_inv(m, pe);
    k = (k * m + 1) / pe;
    ll y = mod_pow(b, k, mod);
    x = bt.mul(x, y);
    return x;
  };

  auto pf = factor(k);
  for (auto&& [p, e]: pf) a = solve_pp(p, e, a);
  return a;
}

ll mod_kth_root_64(ll k, ll a, ll mod) {
  assert(primetest(mod) && 0 <= a && a < mod);
  if (k == 0) return (a == 1 ? 1 : -1);
  if (a == 0) return 0;
  if (mod == 2) return a;
  k %= mod - 1;

  ll g = gcd(k, mod - 1);
  if (mod_pow_64(a, (mod - 1) / g, mod) != 1) return -1;

  ll c = mod_inv(k / g, (mod - 1) / g);
  a = mod_pow_64(a, c, mod);
  k = i128(k) * c % (mod - 1);
  if (k == 0) return 1;

  g = primitive_root_64(mod);

  auto solve_pp = [&](ll p, ll e, ll a) -> ll {
    ll f = 0;
    ll pf = 1;
    while (((mod - 1) / pf) % p == 0) ++f, pf *= p;
    ll m = (mod - 1) / pf;
    /*
    ・位数 Qm の巡回群
    ・a の p^e 乗根をとりたい。持つことは分かっている
    ・a / x^{p^e} = b を維持する。まずは、b が p で割れる回数を増やしていく。
    */
    ll x = 1, b = a, c = f - e; // b ^ {mp^c} = 1

    ll pc = 1;
    FOR(c) pc *= p;
    ll pe = 1;
    FOR(e) pe *= p;
    // 必要ならば原始 p 乗根に関する離散対数問題のセットアップ

    ll G = mod_pow_64(g, (mod - 1) / p, mod);
    ll M = 0;
    ll GM_inv = -1;
    HashMap<ll> MP;
    if (c) {
      while (M * M < p) ++M;
      MP.build(M);
      ll Gpow = 1;
      FOR(m, M) {
        MP[Gpow] = m;
        Gpow = i128(Gpow) * G % mod;
      }
      GM_inv = mod_pow_64(Gpow, mod - 2, mod);
    }

    while (c) {
      /*
      b^{mp^c} = 1 が分かっている。(b/x^{p^e}})^{mp^{c-1}} = 1 にしたい。
      x = g^{p^{f-c-e}*k} として探す。原始 p 乗根 B, G に対する B = G^k に帰着。
      */
      ll B = mod_pow_64(b, pc / p * m, mod);
      ll k = [&](ll B) -> ll {
        FOR(m, M + 1) {
          if (MP.count(B)) return m * M + MP[B];
          B = i128(B) * GM_inv % mod;
        }
        return -1;
      }(B);
      x = i128(x) * mod_pow_64(g, pf / pc / pe * k, mod) % mod;

      ll exp = pf / pc * i128(k) % (mod - 1);
      b = i128(b) * mod_pow_64(g, mod - 1 - exp, mod) % mod;
      --c;
      pc /= p;
    }
    ll k = pe - mod_inv(m, pe);
    k = (i128(k) * m + 1) / pe;
    ll y = mod_pow_64(b, k, mod);
    x = i128(x) * y % mod;
    return x;
  };

  auto pf = factor(k);
  for (auto&& [p, e]: pf) a = solve_pp(p, e, a);
  return a;
}

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