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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> B(N);
    long long ans = 0;
    long long ans2 = 0;
    for (int i = 0; i < N; i++)
    {
        cin >> B[i];
    }
    for (int i = 0; i < N - 1; ++i)
    {
        for (int j = i + 1; j < N; j++)
        {
            if (B[i] < B[j])
            {
                ans++;
            }
        }
    }

    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < N; j++)
        {
            if (B[i] < B[j])
            {
                ans2++;
            }
        }
    }

    long long fact;
    cout << ((ans * K) + ans2 * (K * (K - 1) / 2)) % (long long)(1e9 + 7) << endl;
}