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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;

int main(){
 long long int N,K,count=0,b;
  long long int c;
  cin >>N>>K;
  int hai[N];
  for(int i=0;i<N;i++){
    cin >>hai[i];
  }
  for( int i=0;i<N;i++){
    for( int j=i+1;j<N;j++){
      if(hai[i]>hai[j])count++;
    }
  }
 sort(hai,hai+N);
b=N;
 for(int i=0;i<b-1;i++){
   if(hai[i]==hai[i+1])
    N--;
 }
  c=((K-1)*K-(K-1)*K/2)%1000000007;
  c*=(N-1)*N/2;
  c%=1000000007;
  cout <<(c+count*K)%1000000007;
 


}