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>

void bsort(int a[], int N){//大きい順に並び替える
        int i,j;
        for(i=0;i<N-1;i++){
            for(j= N-1;j>i;j--){
                if(a[j -1] < a[j]){
                    int tmp = a[j];
                    a[j] = a[j-1];
                    a[j-1] = tmp;
                }
            }
        }
    }


int main()
{
  int N;
    long K;
    
    scanf("%d", &N);
    // スペース区切りの整数の入力
    scanf("%ld", &K);
  
  int A[N];
  int s[N];
  int save[N];
  int s2[N];
  int so[N];
  int i,j;
  for(i=0;i<N;i++){
	scanf("%d",&s[i]);
    save[i] = s[i];
    s2[i] = s[i];

  }
  int con =  0;
  for(i=0;i<N-1;i++){
	if(s[i] == s[i+1]){
    	con++;
    }   
  }
  
  if(con == N-1){
   printf("0");
    return 0;
  }
  
  bsort(s2,N);
  
  for(i=0;i<N;i++){
	for(j=0;j<N;j++){
		if(s2[i] == save[j]){
			so[i] = i;
        }
    }
  }
  
 
  int num = 0;
  long ans = 0;
  for(i=0;i<K;i++){
  for(j=0;j<N;j++){
	//ans = ans + (N-(so[i]+1))*(K-j-so[i]);
    ans = ans + (N-(i+1))*(K-j-i);
  }
  }
  
 
  
  ans = ans % 1000000007;

  
  printf("%ld",ans);
    return 0;
}