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;
typedef pair<int,int> P;

#define p_ary(ary,a,b) do { cout << "["; for (int count = (a);count < (b);++count) cout << ary[count] << ((b)-1 == count ? "" : ", "); cout << "]\n"; } while(0)
#define p_map(map,it) do {cout << "{";for (auto (it) = map.begin();;++(it)) {if ((it) == map.end()) {cout << "}\n";break;}else cout << "" << (it)->first << "=>" << (it)->second << ", ";}}while(0)

template<typename T1,typename T2>ostream& operator<<(ostream& os,const pair<T1,T2>& a) {os << "(" << a.first << ", " << a.second << ")";return os;}

const char newl = '\n';

const ll mod = 1000000007;
//const ll mod = 998244353;

const int MAX_N = 500010;
ll fact[MAX_N],fact_inv[MAX_N],inv[MAX_N];

ll pow_mod(ll a,ll b) {
    ll ret;
    if (b < 0) ret = pow_mod(a,mod+b-1);
    else if (b == 0) ret = 1;
    else if (b == 1) ret = a;
    else {
        ll c = pow_mod(a,b/2);
        if (b%2) ret = (c*c)%mod*a%mod;
        else ret = c*c%mod;
    }
    return ret;
}

void init(int x) {
    fact[0] = 1;fact[1] = 1;
    for (int i = 2;i <= x;++i) fact[i] = fact[i-1]*i%mod;
    fact_inv[x] = pow_mod(fact[x],-1);
    for (int i = x;i > 0;--i) fact_inv[i-1] = fact_inv[i]*i%mod;
    for (int i = 1;i <= x;++i) inv[i] = fact_inv[i]*fact[i-1]%mod;
}

ll combi(ll a, ll b) {
    return fact[a]*fact_inv[b]%mod*fact_inv[a-b]%mod;
}

ll permu(ll a, ll b) {
    return fact[a]*fact_inv[a-b]%mod;
}

int main() {
    int n,m;
    cin >> n >> m;
    init(m);
    ll ans = 0;
    for (int i = 0;i < n+1;++i) {
        (ans += (i&1 ? -1 :1)*combi(n,i)*permu(m,i)%mod*permu(m-i,n-i)%mod*permu(m-i,n-i)%mod) %= mod;
    }
    cout << (ans+mod)%mod << endl;
}