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(void){
  long long n,k;
  unsigned long long c=1000000007;;
  scanf("%lld%lld",&n,&k);
  int a[n];
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
  
  long long bb[n][2];   //[0]:右の自分より小さい要素
				  //[1]:左の自分より小さい要素	
  for(int i=0;i<n;i++){
	for(int j=0;j<2;j++){
      bb[i][j]=0;
    }
  }
  
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(i==j)continue;
      if(i>j&&a[i]>a[j])bb[i][1]++;
      if(i<j&&a[i]>a[j])bb[i][0]++;
    }
  }
  for(int i=0;i<n;i++){
  //printf("%d %d %d\n",i+1,bb[i][0],bb[i][1]);
    
  }
  long long r=k*(k+1)/2;
  long long l=k*(k-1)/2;
  //printf("%lld\n%lld\n",r,l);
  unsigned long long sum=0,buf;
  //sum+=bb[0][0]*k+bb[n-1][1]*(k-1);
  for(int i=0;i<n;i++){
    //printf("%d %d %d\n",i+1,bb[i][0],bb[i][1]);
    buf=bb[i][0]*r+bb[i][1]*l;
    if(buf>c)buf%=c;
    //printf("%llu\n",buf);
    sum+=buf;
    //printf("%llu\n",sum);
    if(sum>c)sum%=c;
  }
  
  
  printf("%llu",sum%c);
  return 0;
}