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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long k = sc.nextLong();
		Integer[] a = new Integer[n];
		int mod = 1000000000+7;
//		int[] b = new int[n*k];
		
		long count;
		long counta = 0;
		long countb = 0;
		
		for (int i = 0; i < a.length; i++) {
			a[i] = sc.nextInt();
		}
		
		for (int i = 0; i < a.length-1; i++) {
			for (int j = i+1; j < a.length; j++) {
				if (a[i]>a[j]) {
					counta++;
				}
			}
		}
		
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a.length; j++) {
				if (a[i]>a[j]) {
					countb++;
				}
			}
		}
		
		long ans = ((countb % mod) * ((((long)k * (long)(k-1))/2) % mod));
		ans = (ans + (counta*k)%mod)%mod;
		System.out.println(ans);
		
//		count = ((counta * k)%mod + countb * (k * (k -1) / 2)%mod)%mod;
//		count %= mod;
//		System.out.println(count);
		
		sc.close();
	}
}