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;
#define MOD 1000000007
int main() {
    long long N,K;
    cin >> N >>K;
    vector<long long> a(N);
    for (int i=0;i<N;i++){
        cin >> a.at(i);
    }
    
    long long ans = 0;
    for(int i=0; i<N; i++){
        long long p=0,q=0;
        for(int j=0;j<N;j++){
            if(a.at(i) > a.at(j)){
                p += 1;
                if(i < j){
                    q += 1;
                }
            }
        }
        ans += ((q * K)%MOD +  (p * ( K * (K-1) /2)%MOD)%MOD)%MOD;
        ans = ans%MOD;
    }
    
    cout << ans%MOD << endl;
}