Problem:
Score : 500 points

Problem Statement

Count the pairs of length-N sequences consisting of integers between 1 and M (inclusive), A_1, A_2, \cdots, A_{N} and B_1, B_2, \cdots, B_{N}, that satisfy all of the following conditions:

A_i \neq B_i, for every i such that 1\leq i\leq N.

A_i \neq A_j and B_i \neq B_j, for every (i, j) such that 1\leq i < j\leq N.

Since the count can be enormous, print it modulo (10^9+7).

Constraints

1\leq N \leq M \leq 5\times10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the count modulo (10^9+7).

Sample Input 1

2 2

Sample Output 1

2

A_1=1,A_2=2,B_1=2,B_2=1 and A_1=2,A_2=1,B_1=1,B_2=2 satisfy the conditions.

Sample Input 2

2 3

Sample Output 2

18

Sample Input 3

141421 356237

Sample Output 3

881613484

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
const int INF = 100010001;
const ll LINF = (ll)INF*INF*10;

const int mod = 1000000007;
//const int mod = 998244353;

struct mint {
    int64_t x;
    mint(int64_t x=0):x((x%mod+mod)%mod){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(int64_t t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    //for prime mod
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) {return mint(*this) /= a;}
};

istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

struct combination {
    vector<mint> frac, ifrac;
    combination(int n):frac(n+1), ifrac(n+1) {
        assert(n < mod);
        frac[0] = 1;
        for (int i = 1; i <= n; ++i) frac[i] = frac[i-1]*i;
        ifrac[n] = frac[n].inv();
        for (int i = n; i >= 1; --i) ifrac[i-1] = ifrac[i]*i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return frac[n]*ifrac[k]*ifrac[n-k];
    }
} c(1000000);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    mint ans = c.frac[m]/c.frac[m-n];
    map<P, mint> mp;
    map<P, bool> check;
    function<mint(int, int)> dfs = [&] (int i, int j) {
        if(check[P(i, j)]) return mp[P(i, j)];
        if(!i) return c.frac[j];
        if(i < 0) return mint(0);
        mint res = 0;
        res += mint(i-1)*dfs(i-2, j+1);
        res += mint(j) * dfs(i-1, j);
        check[P(i, j)] = true;
        return mp[P(i, j)] = res;
    };
    cout << ans*dfs(n, m-n) << endl;
}