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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll=long long int;
ll mod=1e9+7;
int main()
{
    ll n,k;
    cin>>n>>k;
    vector<ll> a(n);
    for(int i=0;i<n;i++)
    cin>>a[i];
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        int ct1=0;
        int ct2=0;
        for(int j=0;j<i;j++)
        {
            if(a[j]<a[i])
            ct1++;
        }
        for(int j=i+1;j<n;j++)
        {
            if(a[j]<a[i])
            ct2++;
        }
        ans=(ans+ct1*(k*(k-1))/2)%mod;
        ans=(ans+(ct2*(k*(k+1))/2))%mod;
    }
    cout<<ans<<endl;
    
}