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: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt(), k=sc.nextInt();
		long kei = (long)k*(k-1)/2;
		final long mod = 1000000007;
		int[] array = new int[n];
		for(int i=0;i<n;i++) array[i]=sc.nextInt();
		sc.close();
		long count=0, temp=0;
		for(int i=0;i<n-1;i++) {
			for(int j=i+1;j<n;j++) {
				if(array[i]>array[j]) count++;
			}
		}
		count*=k;
		count%=mod;
		Arrays.sort(array);
		for(int i=n-1;i>=1;i--) {
			temp++;
			if(array[i]!=array[i-1]) {
				count+=(((temp%mod)*i)%mod*kei)%mod;
				temp=0;
			}
			count%=mod;
		}
		System.out.println(count);
	}
}