卡特兰数，组合数，斯特林数，逆元模板

　　直接上板子

const ll mod =(ll) 1e9+7;
const int maxn = 2100001;
ll f[maxn];
int n;

void init() {
    f[0] = f[1] = 1;
    for(int i = 2; i < maxn; i++) f[i] = (f[i-1] * i) % mod;
}
ll mul(ll x, ll n) {
    ll ret = 1;
    while(n) {
        if(n & 1) ret = (ret * x) % mod;
        n >>= 1;
        x = (x * x) % mod;
    }
    return ret;
}
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if(b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    else {
        ll ret = exgcd(b, a%b, x, y);
        ll tmp = x;
        x = y;
        y = tmp - a / b * y;
        return ret;
    }
}
ll inv(ll a) {
    ll x, y;
    exgcd(a, mod, x, y);
    return (x % mod + mod) % mod;
}
ll C(ll x, ll y) {
    return f[x] * inv(f[x-y]) % mod * inv(f[y]) % mod;
}
ll Catalan(int n) {
    return C(2*n, n) * inv(n+1) % mod;
}

ll s[maxn][maxn];
void init2() {
    mem0(s);
    s[1][1] = 1;
    for( int i = 2; i <= maxn-1; i++ ) {
        for( int j = 1; j <= i; j++ ) {
            s[i][j] = s[i-1][j-1]+j*s[i-1][j];
            if( s[i][j] >= mod ) s[i][j] %= mod;
        }
    }
}

ll S(int n, int m) {
    return f[m]*s[n][m]%mod;
}