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;
typedef long long ll;
const ll MOD = 1e9 + 7;
bool vec[2010];
int main(){
ll N,K; cin >> N >> K; vector<ll> A(N);

  for(ll i = 0; i < N; i++){ cin >> A[i]; }
                            
  ll cnt = 0;
  for(ll i = 0; i < N-1; i++){
    for(ll j = i+1; j < N; j++){ if( A[i] > A[j]){ cnt++;}}}
    //A一つあたりの転倒数を求める
   cnt %= MOD;
   cnt *= K; cnt %= MOD; 
  
  sort(A.begin(),A.end());
  ll count = 0;
 
  for(ll i = 0; i < N; i++){ ll res = 0;
    for(ll j = 0; j < N; j++){
      if(A[j] > A[i]){ res++;}}
      count += res; count %= MOD;}
  ll a = (K*(K-1))/2; a %= MOD; count *= a; count %= MOD;
  
  cnt += count; cnt %= MOD;
                           
    
    cout << cnt << endl; return 0;}