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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author tatsumack
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>

#define int long long
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define FOR(i, a, b) for (int i = (a), i##_len = (b); i <= i##_len; ++i)
#define REV(i, a, b) for (int i = (a); i >= (b); --i)
#define CLR(a, b) memset((a), (b), sizeof(a))
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define INF 1001001001001001001ll
#define fcout cout << fixed << setprecision(12)

using namespace std;

typedef pair<int, int> P;

// mod
int mod = 1e9 + 7;

struct mint {
    unsigned x;

    mint() : x(0) {}

    mint(signed sig) { x = sig < 0 ? sig % mod + mod : sig % mod; }

    mint(signed long long sig) { x = sig < 0 ? sig % mod + mod : sig % mod; }

    int get() const { return (int) x; }

    mint& operator+=(mint that) {
        if ((x += that.x) >= mod) x -= mod;
        return *this;
    }

    mint& operator-=(mint that) {
        if ((x += mod - that.x) >= mod) x -= mod;
        return *this;
    }

    mint& operator*=(mint that) {
        x = (unsigned long long) x * that.x % mod;
        return *this;
    }

    mint& operator/=(mint that) { return *this *= that.inverse(); }

    mint operator+(mint that) const { return mint(*this) += that; }

    mint operator-(mint that) const { return mint(*this) -= that; }

    mint operator*(mint that) const { return mint(*this) *= that; }

    mint operator/(mint that) const { return mint(*this) /= that; }

    mint inverse() const {
        long long a = x, b = mod, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b;
            std::swap(a, b);
            u -= t * v;
            std::swap(u, v);
        }
        return mint(u);
    }

    bool operator==(mint that) const { return x == that.x; }

    bool operator!=(mint that) const { return x != that.x; }

    mint operator-() const {
        mint t;
        t.x = x == 0 ? 0 : mod - x;
        return t;
    }
};

struct combination {
    vector<mint> fact, ifact;

    combination(int n) : fact(n + 1), ifact(n + 1) {
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inverse();
        for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }

    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};

struct permutation {
    vector<mint> fact, ifact;

    permutation(int n) : fact(n + 1), ifact(n + 1) {
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inverse();
        for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }

    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[n - k];
    }
};

class ENEQ {
public:
    void solve(std::istream& cin, std::ostream& cout) {
        int N, M;
        cin >> N >> M;

        permutation p(M);
        combination cb(M);

        mint res = 0;
        REP(i, N + 1) {
            mint sign = 1;
            if (i & 1) sign = -1;
            res += sign * p(M, N) * p(M - i, N - i) * cb(N, i);
        }
        cout << res.get() << endl;
    }
};


signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    std::istream& in(std::cin);
    std::ostream& out(std::cout);

    ENEQ solver;
    solver.solve(in, out);
    return 0;
}