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// https://blog.csdn.net/kiwi_berrys/article/details/54834977
// https://www.cnblogs.com/zjp-shadow/p/7773566.html
//阶乘逆元
#define ll long long
ll fac[1000000 + 5]; //阶乘
ll inv[1000000 + 5]; //逆元
void getfac() {
fac[0] = inv[0] = 1;
for (int i = 1; i <= 1000000; i++) {
fac[i] = fac[i - 1] * i % mod;
inv[i] = quick_power(fac[i], mod - 2);
//表示i的阶乘的逆元
}
}
//组合数
inline ll getC(ll n, ll m) // C(n,m) = n!/((n-m)!*m!) % mod
{
return fac[n] * inv[n - m] % mod * inv[m] % mod;
}
//当 nm 过大时可以用Lucas定理降数据
inline ll Lucas(ll n, ll m) {
if (n < mod && m < mod) return getC(n, m);
return Lucas(n / mod, m / mod) * getC(n % mod, m % mod) % mod;
}
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -
//排列数
inline ll
getA(ll n, ll m) // A(n,m) = n!/(n-m)! % mod
{
return n * inv[n - m] % mod;
}