const int mod = 998244353;
const int N = 5e5 + 10;
ll fac[N], ifac[N], inv[N];
void init() {
    fac[0] = ifac[0] = ifac[1] = inv[1] = 1;
    for(int i = 1; i < N; i++) {
        fac[i] = fac[i - 1] * i % mod;
        if(i > 1) ifac[i] = (mod - mod / i) * ifac[mod % i] % mod, inv[i] = ifac[i];
    }
    for(int i = 1; i < N; i++) ifac[i] = ifac[i - 1] * ifac[i] % mod;
}
ll C(int n, int m) {
    return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
ll ksm(ll a, ll b = mod - 2) {
    ll res = 1;
    while(b) {
        if(b & 1) res = res * a % mod;
        a = a * a % mod, b /= 2;
    }
    return res;
} 

使用组合数必须init，预处理O(N) ksm(a)表示求a的倒数（逆元）在mod意义下的值，要求mod为质数