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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
    // Your code here!
    int n,k;
    long all;
    int i;
    int ans=0;
    int spin=0;
    int spent=1;
    int between=0;
    scanf("%d %d",&n, &k);
    int data[n];
    for(i=0;i<n;i++){
        scanf("%d",&data[i]);
        if(i!=0){
            if(data[i-1] >data[i]){
                spin++;
            }
            
        }
    }
    int ter;
    for(i=0;i<n;i++){
        spent=0;
        while(spent<=n){
            ter=i+spent;
            if(ter >=n){
                ter-=n;
            }
            if(data[i] >data[ter]){
                between++;
            }   
            spent++;
        }
       
    }
    ans=spin;
    for(i=0;i<k-1;i++){
        ans=ans+spin+between;
        if(ans >=1000000007){
            ans-=1000000007;
        }
    }
    //ans=((between*(k-1))%1000000007);
    //ans= ans+(spin*k)%1000000007;
    //ans%=1000000007;
    printf("%d\n",ans);
    //printf("%d\n",(between*(k-1)+spin*k)%1000000007);
    //printf("%d %d %d\n",between,spin,k);

    return 0;
}