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: Runtime Error)

#include <iostream>

using namespace std;

int main()
{
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    
    int n,k;
    cin >> n >> k;
    int a[n];
    for(int i=0; i<n; i++){
        cin >> a[i];
    }
    int b[n*k];
    int cnt = 0;
    for(int i=0; i<k; i++){
        for(int j=0; j<n; j++){
            //if(i == 0) cin >> a[i];
            b[cnt++] = a[j];
        }
    }
    cnt = 0;
    for(int i=0; i<n*k; i++){
        //cout << "---" << endl;
        for(int j=0; j<n*k; j++){
            //cout << "B[" << i << "] = " << b[i] << ", B[" << j << "] = " << b[j]; 
            if(i < j){
                if(b[i] > b[j]){
                    cnt++;
                    //cout << " -> OK" << endl;
                }else{
                    //cout << " -> NG" << endl;
                }
            }else{
                //cout << endl;
            }
        }
    }
    
    //cout << "---" << endl;
    cout << cnt%1000000007 << endl;
    return 0;
}