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>
#include<math.h>
#include <string.h>



int main() {
	int i, j, k, l;
	long long int m, h, g;
	long long int f = 0;
	int n;
	int a[2100];
	long long int r1, r2;
	long long int z;
	g = 0;

	scanf("%d %d", &n, &k);
	for (i = 0;i < n;i++) {
		scanf("%d", &a[i]);
	}
	
	r1 = k * (k - 1) / 2;
	r1 = r1 % 1000000007;

	r2 = k * (k + 1) / 2;
	r2 = r2 % 1000000007;

	for (i = 0;i < n;i++) {
		for (j = 0;j < n;j++) {
			if (a[i] < a[j]) {
				g = g + 1;
			}
		}
	}

	z = r1 * g;
	z = z % 1000000007;

	h = 0;
	for (i = 0;i < n;i++) {
		for (j = i;j < n;j++) {
			if (a[i] < a[j]) {
				h = h + 1;
			}
		}
	}

	r2 = r2 * h;
	r2 = r2 % 1000000007;
	z = z + r2;
	z = z % 1000000007;
	
	f = z;

	printf("%lld", f);

	return 0;
}