【模板】两个求逆元的函数

递归：O(logn)

ll inv(ll n)
{
if(n == 1) return 1;
 return ((ll)mod - mod / n) * inv(mod % n, mod) % mod;
}

线性：O(n)

ll inv[maxn];
ll get_inv(ll n)
{
 inv[1] = 1;
 for (ll i = 2; i <= n; i++)
  inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}