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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int m = 1000000007;
int main(){
  ll n,k;
  cin >> n >> k;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  ll inv1=0,inv2=0;
  rep(i,n){
    rep(j,n){
      if(i==j) continue;
      if(a[i]>a[j]){
        if(i<j) inv1++;
        inv2++;
      }
    }
  }
  ll ans=0;
  ans+=(k*inv1)%m;
  ans+=((k*(k-1))/2)%m*inv2;
  ans%=m;
  cout<< ans;
  return 0;
}