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 int inf=1e9;
const int mod=1e9+7;
int main(void)
{
    ll N,K;
    int A[2005];
    ll ans=0;
    cin >> N >> K;
    for (int i = 0; i < N; i++)
    {
        cin >> A[i];
    }
    for(int i=0;i<N;i++)
    {
        ll firstinv=0;
        ll loopinv=0;
        int a=A[i];
        for(int j=i;j<N;j++)
        {
            if(A[j]<a) firstinv++;
        }
        for(int j=0;j<N;j++)
        {
            if(A[j]<a) loopinv++;
        }
        firstinv=(firstinv*K)%mod;
        ans=(ans+firstinv)%mod;
        if(K>=2) 
        {
            ll times=(((K)*(K-1))/2LL)%mod;
            loopinv=(loopinv*times)%mod;
        }
        ans=(ans+loopinv)%mod;
    }
    cout << ans << endl;
    
}