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(){
  long long  n,k,a[2001];
  long long t=0,ans=0,cntr=0,cntl=0,i,j,l,M=1e9+7;
  scanf("%lld%lld",&n,&k);
  for(i=0;i<n;i++){
    scanf("%lld",&a[i]);
    if(i!=0){
      for(j=0;j<i;j++){
        if(a[j]<a[i])cntl++;
      }
    }
  }

for(i=0;i<n;i++){
  for(j=n-1;j>i;j--){
    if(a[j]<a[i])cntr++;
  }
}

/*for(int i=0;i<n;i++){
  printf("%d %d\n",cntr[i],cnt[i]);
}*/
//printf("%lld %lld\n",cntl,cntr );

for(i=1;i<=k;i++){
  t+=((cntr+cntl)*(i-1)+cntr);
}
  printf("%lld\n",t%M);
  return 0;
}