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: Runtime Error)

#include<stdio.h>
int main(){
  int N;
  long int K;
    scanf("%d %ld",&N,&K);
    int A[N];
  int B[N*K];
  for(int i=0;i<N;i++){ 
    scanf("%d",&A[i]);
    for(int k=0;k<K;k++){
      B[i+k*N] = A[i];
    }
  }
  
  long int ans=0;
  for(int i=0;i<N*K;i++){
    for(int j=i;j<N*K;j++){
      if(B[i]>B[j]) ans++;
    }
  }
  
  printf("%ld",ans%1000000009);
  
 return 0; 
}