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;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T = int> void print(vector<T> v){for(auto a:v)cout<<a<<" ";cout<<endl;}

template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) { }
	ModInt(signed sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b; std::swap(a, b);
			u -= t * v; std::swap(u, v);
		}
		if (u < 0) u += Mod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}
	ModInt pow(long long n){
		ModInt r=1,x2=x;
    	while(n){
    		r*=((n%2)?x2:1);
    		x2*=x2;
    		n>>=1;
    	}
    	return r;
	}
};
typedef ModInt<1000000007> mint;
vector<mint> fact, factinv;
void nCr_computeFactinv(int N) {
	N = min(N, mint::Mod - 1);
	fact.resize(N + 1); factinv.resize(N + 1);
	fact[0] = 1;
	reg(i, 1, N) fact[i] = fact[i - 1] * i;
	factinv[N] = fact[N].inverse();
	for (int i = N; i >= 1; i --) factinv[i - 1] = factinv[i] * i;
}
mint nCr(int n, int r) {
	if (n >= mint::Mod)
		return nCr(n % mint::Mod, r % mint::Mod) * nCr(n / mint::Mod, r / mint::Mod);
	return r > n ? 0 : fact[n] * factinv[n - r] * factinv[r];
}
mint nPr(int n, int r) {
	if (r<0)return 0;
	return r > n ? 0 : fact[n] * factinv[n-r];
}
mint nHr(int n, int r) { return r == 0 ? 1 : nCr(n + r - 1, r); }

/*
異なるみたいなやつは、包除原理でいける
*/

ll n,m;
mint ans;

int main(void){
	cin>>n>>m;
	nCr_computeFactinv(500010);
	reg(i,0,n){
		ans += nPr(m,i)*nPr(m-i,n-i)*nPr(m-i,n-i)*nCr(n,i) * (i%2==0?1:-1);
	}
	cout<<ans.get()<<endl;
	return 0;
}