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>
#define int long long
#define INT_MAX INT64_MAX
#define INT_MIN INT64_MIN
#define PI 3.14159265358979323846
#define CLK clock_t clk = clock();//Start of main
#define OCLK cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / CLOCKS_PER_SEC << '\n';//End of main
#define MOD 1000000007
/* DEFINE cout */
#define pln cout << "===============================\n"
#define on cout<<"\n"
#define os cout<<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define bitcount __builtin_popcount
/* DEFINE cout */
#define endl "\n"
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);//fastio
using namespace std;
struct hash_pair { // use pair as key in unordered_map<pair<int,int>,int,hash_pair>;
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const
    {
        auto hash1 = hash<T1>{}(p.first);
        auto hash2 = hash<T2>{}(p.second);
        return hash1 ^ hash2;
    }
};
bool prime[10000001];
int P=MOD-2;
int factorial[10000001]={0};
int powerFunction(int x,int y){
    int res = 1;
    int p = MOD;
    x = x % p;
    while (y > 0){
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
void sieveFunction(int maxLimit){
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1]=false;
    for(int i=2; maxLimit >= i ; i++){
        if(prime[i]){
            for(int j=2*i ; maxLimit >= j ; j+=i)
                prime[j]=false;
        }
    }
}
void factorialFunction(int maxLimit){
    factorial[0]=1;
    for(int i=1 ; i <= maxLimit ; i++)
        factorial[i]=(factorial[i-1]*i)%MOD;
    return;
}
int gcd(int a, int b){
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
int ncr(int n, int r){
    if(r > n || r < 0){
        return 0;
    }
    int temp = factorial[n]%MOD;
    temp *= powerFunction(factorial[r], MOD-2)%MOD;
    temp %= MOD;
    temp *= powerFunction(factorial[n-r], MOD-2)%MOD;
    temp %= MOD;
    return temp;
}

signed main(){
    int n, m;
    cin >> n >> m;
    factorialFunction(500005);
    int del = 0;
    for(int i=1; n>=i; i++){
        int temp = ((ncr(n, i)%MOD)*(ncr(m, i)%MOD))%MOD;
        temp *= factorial[i];
        temp %= MOD;
        int x = ncr(m-i, n-i)%MOD;
        x *= factorial[n-i]%MOD;
        x %= MOD;
        x *= x;
        x %= MOD;
        temp *= x;
        temp %= MOD;
        if(i%2 == 1){
            del += temp;
            del %= MOD;
        }else{
            del += MOD;
            del -= temp;
            del %= MOD;
        }
    }
    int total = ncr(m, n)%MOD;
    total *= factorial[n];
    total %= MOD;
    total *= total;
    total %= MOD;
    int ans = total;
    ans += MOD;
    ans -= del;
    ans %= MOD;
    cout << ans << endl;
}