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 <string>
using namespace std;

#define W (1000000000+7)

int main()
{
	unsigned long int n,k,a[2001],i,j,b[2001],c1=0,c2,c3,c4;
	
	cin>>n>>k;
	for(i=0;i<n;i++)
		cin>>a[i];
	
	for(i=0;i<n;i++)
		for(j=i+1;j<n;j++)
		{
			if(a[i]>a[j])
				b[i]++;
		}
	
	for(i=0;i<n;i++)
		c1+=b[i];
	
	c2=c1;
	
	for(i=0;i<n;i++)
		for(j=i+1;j<n;j++)
		{
			if(a[i]<a[j])
				b[j]++;
		}
	
	//cout<<k<<" "<<k*(k-1)/2*c2<<endl;
	
	//cout<<((k*(k-1)/2)*c2+c1*k)%(1000000000+7);
	if(k%2)
	{
		c3=(k-1)/2;
		c3=c3*k%W;
	}
	else
	{
		c3=k/2;
		c3=c3*(k-1)%W;
	}
	c3=c3*c2%W;
	
	c1=c1*k%W;
	
	cout<<(c1+c3)%W;
	
	
	return 0;
}