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)

#ifdef __APPLE__
#include <iostream>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <iomanip>
#include <cmath>
#include <ctgmath>
#include <math.h>
#include <cassert>
#include <chrono>
#else
#include <bits/stdc++.h>
#endif

using namespace std;
#define io ios_base::sync_with_stdio(false), cin.tie(NULL)
#define GFOR(i,a,n) for (i=a;i<n;i++)
#define FOR(i,a,n) for (int i=a;i<n;i++)
#define RFOR(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
#define tc(t) for (int tc=1; tc<=t; tc++)
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<long long int> vll;
typedef vector<pair<int, int> > vpi;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
mt19937 mrand(random_device{}()); 
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll modInverse (ll a) {ll p = mod - 2; return powmod(a, p);}
// head
vll fac(500001);

void init() {
    fac[0] = fac[1] = 1;
    FOR (i, 2, sz(fac))
        fac[i] = (i*fac[i-1])%mod;
}

ll ways(ll n, ll k) {
    return fac[n]*((modInverse(fac[k])*modInverse(fac[n-k]))%mod)%mod;
}

void solve() {
    init();

    ll n, m;
    cin >> n >> m;

    ll ans = 0, temp;
    FOR (i, 0, n+1) {
        temp = (fac[n-i]*ways(m-i, n-i))%mod;
        temp = (temp*temp)%mod;
        temp = (((((fac[i]*ways(n, i))%mod)*ways(m, i))%mod)*temp)%mod;
        if (i%2 == 0) {
            ans += temp;
        } else {
            ans -= temp;
        }
        ans = (ans + mod)%mod;
    }

    cout << ans << "\n";
}

int main() {
    #ifdef LOCAL
    auto t1 = std::chrono::high_resolution_clock::now();
    #endif

    io;
    int t = 1;
    tc(t) {
        solve();
    }

    #ifdef LOCAL
    auto t2 = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::milliseconds>( t2 - t1 ).count();

    std::cout << duration << "\n";
    #endif

}
//check for edge cases in limits
//printing double may print 1.223232e9 like values, beware of that