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>
#include <cstdio>
using namespace std;
using ll = long long;
const ll mod = 1e9+7;
int main() 
{
    int n, k;
    cin >> n >> k;
    int a[2010] = {};
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }

    int cntl[2010] = {};
    int cntr[2010] = {};
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < i; j++)
        {
            if (a[i] < a[j]) cntl[i]++;
        }
        for (int j = i+1; j < n; j++)
        {
            if (a[i] < a[j]) cntr[i]++;
        }
        
    }

    ll ans = 0;
   for (int i = 0; i < n; i++)
        {
            ans = (ans + (ll)k*(2*cntl[i])/2%mod+((ll)k*(ll)(k-1))%mod/2*(cntl[i]+cntr[i])%mod) % mod;
            //ans = (ans + k%mod*(((k-1)%mod*(cntl[i]+cntrs[i]))/2+cntl[i])%mod) % mod;
            //ans = (ans + (ll)k * cntl[i] % mod  + ((ll)k*((ll)k-1)/2) % mod * (cntl[i]+cntr[i]))%mod;
            //ans = (ans + (ll)k * cntl[i] % mod  + ((ll)k*(ll)(k-1)/2) % mod * (cntl[i]+cntr[i]))%mod;
        }
    
    
    cout << ans << "\n";
    
    
    
    
}