Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;

namespace B___Kleene_Inversion
{
    class Program
    {
        static void Main(string[] args)
        {
            var NK = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
            var N = NK[0];
            var K = NK[1];
            var A = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);

            var invert = new long[N];
            for (int i = 0; i < N; i++) for (int j = i + 1; j < N; j++) if (A[i] > A[j]) invert[i]++;

            var lessCount = new long[N];
            for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) if (i != j && A[i] > A[j]) lessCount[i]++;

            var result = new ModInt();
            for (int i = 0; i < N; i++)
            {
                result += (long)(invert[i] * K);
                result += (long)(lessCount[i] * (K * K - K) / 2);
            }
            Console.WriteLine(result);
        }
    }
    //パクリ元　https://atcoder.jp/contests/arc067/submissions/5337727
    struct ModInt
    {
        const int MOD = 1000000007;
        long Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public ModInt(long data) { if ((Data = data % MOD) < 0) Data += MOD; }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator long(ModInt modInt) => modInt.Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator ModInt(long val) => new ModInt(val);
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, ModInt b)
        {
            long res = a.Data + b.Data;
            return new ModInt() { Data = res >= MOD ? res - MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, long b) => a.Data + b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, ModInt b)
        {
            long res = a.Data - b.Data;
            return new ModInt() { Data = res < 0 ? res + MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, long b) => a.Data - b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator *(ModInt a, ModInt b) => new ModInt() { Data = a.Data * b.Data % MOD };
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator /(ModInt a, ModInt b) => a.Data * GetInverse(b);

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public override string ToString() => Data.ToString();

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        static long GetInverse(long a)
        {
            long div;
            long p = MOD;
            long x1 = 1, y1 = 0, x2 = 0, y2 = 1;
            while (true)
            {
                if (p == 1) return x2;
                div = a / p;
                x1 -= x2 * div;
                y1 -= y2 * div;
                a %= p;
                if (a == 1) return x1;
                div = p / a;
                x2 -= x1 * div;
                y2 -= y1 * div;
                p %= a;
            }
        }
    }
}