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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>

int main()
{
    int n;
    int k;
    int cnt = 0;
    int sum = 0;
    
    scanf("%d", &n);
    scanf("%d", &k);
    
    int a[n];
    for (int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < n; i++){
        int pvt = a[i];
        for (int j = i; j < n; j++){
            if (a[j] < pvt) cnt++;
        }
    }
    
    sum = cnt * k;
    int cnt_k = 0;
    for (int i = 0; i < n; i++){
        int pvt = a[i];
        for (int j = 0; j < n; j++){
            if (a[j] < pvt) cnt_k++;
        }
    }
    
    sum += (cnt_k * ((k*k-1)/2));
    printf("%d\n", sum%1000000007);
    return 0;

}