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)

#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <random>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define repp(i,n,m) for (int i = m; i < (n); ++i)
#define repl(i,n) for (long long i = 0; i < (n); ++i)
#define reppl(i,n,m) for (long long i = m; i < (n); ++i)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using PL = pair<long long, long long>;
using Pxy = pair<double, double>;
using T = tuple<int, int, bool>;
const int INF = 1001001007;
const int mod = 1000000007;
const int MOD = 998244353;
const ll inf = 1e18;
template <typename AT>
void printvec(vector<AT> &ar){
    rep(i,ar.size()-1) cout << ar[i] << " ";
    cout << ar[ar.size()-1] << endl;
}
template <typename Q>
void printvvec(vector<vector<Q>> &ar){
    rep(i,ar.size()){
        rep(j,ar[0].size()-1) cout << ar[i][j] << " ";
        cout << ar[i][ar[0].size()-1] << endl;
    }
}
template <typename S>
bool range(S a, S b, S x){
    return (a <= x && x < b);
}

struct mint {
  ll x; // typedef long long ll;
  mint(ll 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 {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll 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) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(int n) {
  if (n == 0) return 1;
  mint x = f(n/2);
  x *= x;
  if (n%2 == 1) x *= 2;
  return x;
}

mint choose(int n, int a) {
  mint x = 1, y = 1;
  rep(i,a) {
    x *= n-i;
    y *= i+1;
  }
  return x / y;
}

ll gcd (ll a, ll b){
    ll k = a % b;
    while (k > 0){
        a = b;
        b = k;
        k = a % b;
    }
    return b;
}

int main() {
    int n, m; cin >> n >> m;
    if (n == 1){
        cout << m * m - m << endl;
        return 0;
    }
    if (n > m){
        cout << 0 << endl;
        return 0;
    }
    cout << 1000000007 << endl;
}