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 <iostream>
#include <vector>
#include <string>
using namespace std;
using ll = long long;

int main(){
    ll n,k; cin >> n >> k;
    ll a[n+1],b[3*n+2];
    ll c[3*n+2];

    ll res = 0;
    for(int i=0;i<n;i++){
        cin >> a[i];
        b[n-1-i] = a[i];
        b[2*n-1-i] = a[i];
        b[3*n-1-i] = a[i];
      
    }
    for(int i=0;i<3*n;i++){
        c[i]=0;
        for(int j=i;j<3*n;j++){
            if(b[i] < b[j]) c[i]++;
        }
    }
    ll d = 0; ll st = 0;
    for(int i=0;i<n;i++){
      st += c[i+2*n];
      d += c[i+n]-c[i+2*n];
    } 
    
    ll ans = k*(((2*st+(k-1)*d)/2)% 1000000007) % 1000000007;
    cout << ans;
    return 0;
}