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],array2[length]; //２が操作用
    ll a[5];
    int *inp;
    int N,i,j,temp,m,ttt,N2;
    ll sum=0,K,v,sumtemp=0;
    scanf("%d%lld",&N,&K);
    for(i=0;i<N;i++){
      scanf("%d",array+i);
      *(array2+i)=*(array+i);
    }
    qsort(array2,length,sizeof(int),(int(*)(const int*, const int*))compare);
   N2=N/2;
    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)>*(array2+j)){
            temp++;}
            else{
                break;
            }
        }
        sumtemp+= temp;
        for(m=i;m<N;m++){
            if(*(array+i)>*(array+m)){
                ttt++;
            }
            sum= sum + ttt*K;
        }
        if(i==(N/5)){
           a[0]= sum+sumtemp*(sumtemp*K*(K-1)/2);
           sum=0; 
           sumtemp=0;
        }
        if(i==(N*2/5)){
           a[1]= sum+sumtemp*(sumtemp*K*(K-1)/2);
           sum=0; 
           sumtemp=0;
        }
        if(i==(N*3/5)){
           a[2]= sum+sumtemp*(sumtemp*K*(K-1)/2);
           sum=0; 
           sumtemp=0;
        }
        if(i==(N*4/5)){
           a[3]= sum+sumtemp*(sumtemp*K*(K-1)/2);
           sum=0; 
           sumtemp=0;
        }
    }
        sum=sum%1000000007;
        sumtemp=sumtemp%1000000007;
        sum= sum+sumtemp;
        sum+=a[0]%1000000007+a[1]%1000000007+a[2]%1000000007+a[3]%1000000007;
    printf("%lld",sum%1000000007);
    return 0;
}