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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using CompLib.Mathematics;
using CompLib.Util;

public class Program
{
    private int N, M;

    private ModInt[] F;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        M = sc.NextInt();
        F = new ModInt[1000001];
        F[0] = 1;
        for (int i = 1; i <= 1000000; i++)
        {
            F[i] = F[i - 1] * i;
        }

        // Aの選び方 M P N
        var p = P(M, N);
        // Bの選び方

        ModInt q = 0;
        for (int i = 0; i <= N; i++)
        {
            // i箇所一致させる
            // のこり N-iには適当
            var e = P(M - i, N - i) * C(N, i);
            if (i % 2 == 0) q += e;
            else q -= e;
        }

        Console.WriteLine(p * q);
    }

    ModInt C(int n, int r)
    {
        return F[n] * ModInt.Inverse(F[n - r] * F[r]);
    }

    ModInt P(int n, int r)
    {
        return F[n] * ModInt.Inverse(F[n - r]);
    }

    public static void Main(string[] args) => new Program().Solve();
}

namespace CompLib.Util
{
    using System;
    using System.Linq;

    class Scanner
    {
        private string[] _line;
        private int _index;
        private const char Separator = ' ';

        public Scanner()
        {
            _line = new string[0];
            _index = 0;
        }

        public string Next()
        {
            while (_index >= _line.Length)
            {
                _line = Console.ReadLine().Split(Separator);
                _index = 0;
            }

            return _line[_index++];
        }

        public int NextInt() => int.Parse(Next());
        public long NextLong() => long.Parse(Next());
        public double NextDouble() => double.Parse(Next());
        public decimal NextDecimal() => decimal.Parse(Next());
        public char NextChar() => Next()[0];
        public char[] NextCharArray() => Next().ToCharArray();

        public string[] Array()
        {
            _line = Console.ReadLine().Split(Separator);
            _index = _line.Length;
            return _line;
        }

        public int[] IntArray() => Array().Select(int.Parse).ToArray();
        public long[] LongArray() => Array().Select(long.Parse).ToArray();
        public double[] DoubleArray() => Array().Select(double.Parse).ToArray();
        public decimal[] DecimalArray() => Array().Select(decimal.Parse).ToArray();
    }
}

// https://bitbucket.org/camypaper/complib
namespace CompLib.Mathematics
{
    #region ModInt

    /// <summary>
    /// [0,<see cref="Mod"/>) までの値を取るような数
    /// </summary>
    public struct ModInt
    {
        /// <summary>
        /// 剰余を取る値．
        /// </summary>
        public const long Mod = (int) 1e9 + 7;

        /// <summary>
        /// 実際の数値．
        /// </summary>
        public long num;

        /// <summary>
        /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
        /// </summary>
        /// <param name="n">インスタンスが持つ値</param>
        /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
        public ModInt(long n)
        {
            num = n;
        }

        /// <summary>
        /// このインスタンスの数値を文字列に変換します．
        /// </summary>
        /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
        public override string ToString()
        {
            return num.ToString();
        }

        public static ModInt operator +(ModInt l, ModInt r)
        {
            l.num += r.num;
            if (l.num >= Mod) l.num -= Mod;
            return l;
        }

        public static ModInt operator -(ModInt l, ModInt r)
        {
            l.num -= r.num;
            if (l.num < 0) l.num += Mod;
            return l;
        }

        public static ModInt operator *(ModInt l, ModInt r)
        {
            return new ModInt(l.num * r.num % Mod);
        }

        public static implicit operator ModInt(long n)
        {
            n %= Mod;
            if (n < 0) n += Mod;
            return new ModInt(n);
        }

        /// <summary>
        /// 与えられた 2 つの数値からべき剰余を計算します．
        /// </summary>
        /// <param name="v">べき乗の底</param>
        /// <param name="k">べき指数</param>
        /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
        public static ModInt Pow(ModInt v, long k)
        {
            return Pow(v.num, k);
        }

        /// <summary>
        /// 与えられた 2 つの数値からべき剰余を計算します．
        /// </summary>
        /// <param name="v">べき乗の底</param>
        /// <param name="k">べき指数</param>
        /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
        public static ModInt Pow(long v, long k)
        {
            long ret = 1;
            for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
                if ((k & 1) == 1)
                    ret = ret * v % Mod;
            return new ModInt(ret);
        }

        /// <summary>
        /// 与えられた数の逆元を計算します．
        /// </summary>
        /// <param name="v">逆元を取る対象となる数</param>
        /// <returns>逆元となるような値</returns>
        /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
        public static ModInt Inverse(ModInt v)
        {
            return Pow(v, Mod - 2);
        }
    }

    #endregion

    #region Binomial Coefficient

    public class BinomialCoefficient
    {
        public ModInt[] fact, ifact;

        public BinomialCoefficient(int n)
        {
            fact = new ModInt[n + 1];
            ifact = new ModInt[n + 1];
            fact[0] = 1;
            for (int i = 1; i <= n; i++)
                fact[i] = fact[i - 1] * i;
            ifact[n] = ModInt.Inverse(fact[n]);
            for (int i = n - 1; i >= 0; i--)
                ifact[i] = ifact[i + 1] * (i + 1);
            ifact[0] = ifact[1];
        }

        public ModInt this[int n, int r]
        {
            get
            {
                if (n < 0 || n >= fact.Length || r < 0 || r > n) return 0;
                return fact[n] * ifact[n - r] * ifact[r];
            }
        }

        public ModInt RepeatedCombination(int n, int k)
        {
            if (k == 0) return 1;
            return this[n + k - 1, k];
        }
    }

    #endregion
}