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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
	long int i, j;
	long int n, k;
	long int sum;
	long int max;
	long int count;
	long int flg;
	long int a[2000];
	
	scanf("%ld%ld", &n, &k);
	
	for(i = 0; i < n; i++) {
		scanf("%ld", &a[i]);
	}
	
	sum = 0;
	for(i = 0; i < n; i++) {
		for(j = 0; j < n; j++) {
			if(i != j) {
				if(a[i] < a[j]) {
					sum++;
				}
			}
		}
	}
	
	sum %= 1000000007;
	sum *= ( ( (k - 1) * k / 2) % 1000000007);
	sum %= 1000000007;
	
	count = 0;
	for(i = 0; i < n - 1; i++) {
		for(j = i + 1; j < n; j++) {
			if(a[i] > a[j]) {
				count++;
			}
		}
	}
	
	sum += ( (count * k) % 1000000007);
	sum %= 1000000007;
	
	printf("%ld\n", sum);
	
	return 0;
}


//185297239
//185297239
//845502410
//9223372036854775807
//19桁