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>
#define MOD 1000000007;


int main()
{   long long n,k;
    scanf("%lld%lld",&n,&k);
    long long a[2001];
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    long long front=0,rear=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<i;j++){
            if(a[i]<a[j]){
                front++;
                //printf("b[%d].f increased @%d\n",i,j);
            }
        }for(int j=i+1;j<n;j++){
            if(a[i]<a[j]){
                rear++;
                //printf("b[%d].r increased @%d\n",i,j);
            }
        }
    }
    long long ans=0;
  
        ans+=(rear*(k+1)*k/2)%MOD;
        ans%=MOD;
        ans+=(front*(k-1)*k/2)%MOD;
        ans%=MOD;
    
    

    
    printf("%lld",ans);

    return 0;
}