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 <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
 
int main() {
    long long n,k,i,ans,coleft,cof,j;
    long long mod=1000000007;
    coleft=0;
    cof=0;
    ans=0;
    cin>>n>>k;
    vector<int> vec(n);
    for(i=0;i<n;i++){
      cin>>vec[i];
    }
    for(i=0;i<n;i++){
      for(j=0;j<i;j++){
        if(vec[i]>vec[j]){
          coleft++;
        }
      }
      for(j=i+1;j<n;j++){
        if(vec[i]>vec[j]){
          cof++;
        }
      }
      ans+=coleft*((k-1)*k/2%mod)%mod;
      coleft=0;
      ans+=cof*(k*(k+1)/2%mod)%mod;
      cof=0;
    }
  ans=ans%mod;
  cout<<ans<<endl;
}