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, a[2000], all[2000], right[2000];
	long long int k, sum;
	scanf("%d %lld", &n, &k);
	int i, j, left, count;
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);

	for (i = 0; i < n; i++)
	{
		count = 0;
		for (j = 0; j < i; j++)
		{
			if (a[j] < a[i])
				count++;
		}
		left = count;
		for (j = i + 1; j < n; j++)
		{
			if (a[j] < a[i])
				count++;
		}
		all[i] = count;
		right[i] = count - left;
	}

	sum = 0;
	for (i = 0; i < n; i++)
	{
		sum = (sum + (k * right[i]) % mod + (((k * all[i]) % mod * (k - 1)) / 2)) % mod;
	}

	printf("%lld\n", sum);
	return 0;
}