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 <iostream>
#include <vector>
#define MOD 1000000007
using namespace std;

int main()
{
    int n, k;
    cin >> n >> k;
    vector<int> A;
    for (int i = 0; i < n; i++)
    {
        int num;
        cin >> num;
        A.push_back(num);
    }

    long long counter1 = 0;
    // セットA内の転倒数を計算する
    for (int i = 0; i < A.size(); i++)
    {
        for (int j = i + 1; j < A.size(); j++)
        {
            if (A[i] > A[j])
            {
                counter1++;
                counter1 = counter1 % MOD;
            }
        }
    }

    long long counter2 = 0;
    for (int i = 0; i < A.size(); i++)
    {
        for (int j = 0; j < A.size(); j++)
        {
            if (A[i] > A[j] && i != j)
            {
                counter2++;
                counter2 = counter2 % MOD;
            }
        }
    }

    long long answer = 0;
    long long keisu = k * (k - 1);
    long long keisu2 = keisu / 2;
    answer += counter1 * k;
    answer = answer % MOD;
    answer += counter2 * keisu2;
    cout << (answer % MOD) << endl;
}