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>
#include<string.h> 
typedef  long long ll;
#define length 3000 //10万
#define kw_len 10+2
int compare(const int*a,const int *b){
    if(*a<*b)
      return -1;
    else if(*a>*b)
      return 1;
    return 0;
}
 
ll nCr(int n,int r){
  ll val;
  int i;
  if(r==0)return 1;
  for(i=0;i<r;i++){
		val *= (n-i);
		val /= (i+1);
	}
  return val;
}
 
ll nPr(int n,int r){
  ll val=1;
  int i;
  for(i=0;i<r;i++){
    val *= (n-i);
  }
  return val;
}
/*
qsort(array,length,sizeof(int),(int(*)(const int*, const int*))compare);
使う時これをコピペ
*/


int main(){
    int array[length]; 
    int N,i,j,m;
    ll sum=0,K,v=0,sumtemp=0,mod=0,temp,ttt;
    scanf("%d%lld",&N,&K);
    for(i=0;i<N;i++){
      scanf("%d",array+i);
    }

    for(i=0;i<N;i++){
        temp=0;
        ttt=0;
        for(j=0;j<N;j++){    //全部でデカいの何個あるか
            if(*(array+i)>*(array+j)){
            temp++;}
        }
        sumtemp= (sumtemp+(temp*K*(K-1)/2))%1000000007;
        for(m=i+1;m<N;m++){
            if(m>=N)break;
            if(*(array+i)>*(array+m)){
                ttt++;
            }
        }
        sum= (sum + ttt*K)%1000000007;
        
    }
        sum= sum+sumtemp;
    printf("%lld",sum%1000000007);
    return 0;
}