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 n,k;
  cin >> n >> k;
  long a[n];
  for(long i = 0;i < n;i++){
    cin >> a[i];
  }
  vector<long> tennum(n),tennum2(n);
  for(long i = 0;i < n;i++){
    for(long j = i+1;j < n;j++){
      if(a[i] > a[j]){
        tennum.at(i)++;
      }
    }
  }
  for(long i = 0;i < n;i++){
    for(long j = 0;j < n;j++){
      if(i == j) continue;
      if(a[i] > a[j]){
        tennum2.at(i)++;
      }
    }
  }
  long ans = 0,inf = 1000000007;
  for(long i = 0;i < n;i++){
    long c = (k*tennum2.at(i)+tennum.at(i))%inf;
    long d = k*(k+1)%inf;
    ans += (c*k - (tennum2.at(i)*d)/2)%inf;
    ans += inf;
    ans %= inf; 
  }
  cout << ans << endl;
}