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: Accepted)

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;


int main(void){
  ll n,k,i,j,a[2003],me[2003]={},mv[2003]={},ans=0;
  const int mod=1e9+7;
  
  cin >> n >> k;
  for (i=0;i<n;i++) cin >> a[i];
  
  for (i=0;i<n-1;i++) for (j=i+1;j<n;j++){
    if (a[i]>a[j]) me[i]++;
  }
  
  for (i=0;i<n;i++) for (j=0;j<n;j++){
    if (i==j) continue;
    if (a[i]>a[j]) mv[i]++;
  }
  
  for (i=0;i<n;i++){
    ans=(ans+k*(k-1)/2%mod*mv[i]%mod+me[i]*k%mod)%mod;
  }
  
  cout << ans << endl;
  
  return 0;
}