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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

#define mod 1000000007
int main(void)
{
	int N,K;
	scanf("%d %d",&N,&K);
	int A[N],i,j;
	int castle[N],kate[N];
	for(i=0;i<N;i++){
		scanf("%d",&A[i]);
		castle[i] = 0;
		kate[i] = 0;
	}
	for(i=0;i<N;i++){
		for(j=0;j<N;j++)
		{
			if(A[i] > A[j]){
				if(i<j) castle[i]++;
				else kate[i]++;
			}
		}
	}
	long long int ans = 0;
	/*
	for(i=0;i<N;i++){
		printf("castle[%d] = %d, kate[%d] = %d\n",i,castle[i],i,kate[i]);
	}
	*/
	for(i=0;i<N;i++){
		ans += castle[i]*K*(K+1)/2;
		ans = ans % mod;
		ans += kate[i]*K*(K-1)/2;
		ans = ans % mod;
	}
	printf("%lld",ans);
	
	return 0;
}