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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
#define REP(i,n) for (ll i = 0; i < (n); ++i)
#define RREP(i, n) for (ll i = (n) - 1; i >= 0; --i)
#define ALL(v) (v).begin(), (v).end()
template<class T>
using reverse_priority_queue = priority_queue<T, vector<T>, greater<T>>;

struct mint{
    static const long long MOD = 1000000007;

    long long v;

    mint() : v(0) {}
    mint(long long x){
        x = x % MOD;
        if(x < 0){ x += MOD; }
        v = x;
    }
    mint& operator+=(mint a){
        v += a.v;
        if(v >= MOD){ v -= MOD; }
        return *this;
    }
    mint& operator*=(mint a){
        v *= a.v;
        v %= MOD;
        return *this;
    }
    mint& operator-=(mint a){
        v -= a.v;
        if(v < 0){ v += MOD; }
        return *this;
    }
    mint& operator/=(mint a){
        return (*this) *= a.inv();
    }

    mint operator*(mint a) const{
        return mint(v) *= a;
    }

    mint operator+(mint a) const {
        return mint(v) += a;
    }

    mint operator-(mint a) const {
        return mint(v) -= a;
    }

    mint operator/(mint a) const {
        return mint(v) /= a;
    }

    mint pow(long long k) const {
        mint res(1),tmp(v);
        while(k){
            if(k&1) res *= tmp;
            tmp *= tmp;
            k >>= 1;
        }
        return res;
    }
    mint inv(){return pow(MOD - 2); }

    static mint comb(long long n, int k){
        mint res(1);
        for(int i = 0; i < k; ++i){
            res *= mint(n - i);
            res /= mint(i + 1);
        }
        return res;
    }

    static mint factorial(long long n){
        mint res(1);
        for(int i = n; i > 1; --i){
            res *= mint(i);
        }
        return res;
    }

    bool operator<(const mint &a) const {
        return v < a.v;
    };
};

mint operator*(long long l, mint r){ return r * l; }

std::ostream& operator<<(std::ostream& stream, const mint& m){
    stream << m.v;
    return stream;
}

struct mintcomb{
    vector<mint> fact;
    vector<mint> ifact;
    mintcomb(ll n) : fact(n + 1, 1), ifact(n + 1, 1) {
        for(ll i = 2; i <= n; ++i){
            fact.at(i) = fact.at(i - 1) * i;
            ifact.at(i) = fact.at(i).inv();
        }
    }

    mint factorial(ll k) const { return fact.at(k); }
    mint permutation(ll k, ll l) const {
        return fact.at(k) * ifact.at(k - l);
    }
    mint combination(ll k, ll l) const {
        return fact.at(k) * ifact.at(l) * ifact.at(k - l);
    }
};

int main(){
    ll n, m;
    cin >> n >> m;
    mintcomb mc(m);
    mint ans = 0;
    REP(k, n + 1){
        ans += (k % 2 == 0 ? 1 : -1) * mc.combination(n, k) * mc.permutation(m - k, n - k); 
    }
    ans *= mc.permutation(m, n);
    cout << ans << endl;
    return 0;
}