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>
#include<stdlib.h>
#include<math.h>

int main(){
    long N,K,*A,i,j,cnt=0,cnt1=0,cnt2=0,l;
    scanf("%ld %ld",&N,&K);
    A=(long*)malloc(N*sizeof(long));
    if(A==NULL){
        printf("error\n");
        return 1;
    }
    for(i=0;i<N;i++){
        scanf("%ld",&A[i]);
    }
    for(i=0;i<N;i++){
            for(j=i+1;j<N;j++){
            if(A[i]>A[j]){
                cnt1++;
            }
        }
        for(j=0;j<N;j++){
            if(A[i]>A[j]){
                cnt2++;
            }
        }
    }
    for(l=1;l<=K;l++){
        cnt+=cnt1+cnt2*(K-l);
    }
    printf("%ld\n",cnt%(1000000000+7));
    free(A);
    return 0;
}