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: Accepted)

#include<stdio.h>
int main()
{
	long long int n, k;
	scanf("%lld %lld", &n, &k);
	long long int i, j;
	int a[2003];
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	long long int p = 1e9 + 7;
	long long int ans = 0;
	long long int c, d;
	for (i = 0; i < n; i++)
	{
		c = 0;
		for (j = i + 1; j < n; j++)
			if (a[j] < a[i])
				c++;
		d = 0;
		for (j = 0; j < n; j++)
			if (a[j] < a[i])
				d++;
		ans += c * k % p + d * ((k - 1) * k / 2 % p) % p;
		ans %= p;
	}
	printf("%lld\n", ans);
	return 0;
}