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;
typedef long long ll;
ll x=1e9+7;
ll a[20000],bq[20000],bl[20000];
int main()
{
	ll n,k;
	cin>>n>>k;
	for(ll i=0;i<n;i++)
	{
		cin>>a[i];
	}
	for(int i=0;i<n;i++)
	{
		ll sum,sum1=0,sum2=0;
		sum=a[i];
		for(ll j=0;j<n;j++)
		{
			if(a[j]<sum&&i>j)
			sum1++;
			else if(a[j]<sum&&i<j)
			sum2++;
		}
		bq[i]=sum1;
		bl[i]=sum2;
	}
	long long sum3=0;
	for(ll i=0;i<n;i++)
	{
		
		sum3=sum3+bq[i]*(k-1)*k/2+bl[i]*k*(k+1)/2;
		sum3=sum3%x;
	}
	cout<<sum3<<endl;
 }