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: Runtime Error)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int a[] = new int[n];
        int count  = 0;
        for(int i = 0; i < n; i++) {
        	a[i] = sc.nextInt();
        }
        int a1[] = new int[n*k];
        for(int i = 0; i < n*k; i += n) {
        	for(int j = 0; j < n; j++) {
            	a1[i+j] = a[j];
            }
        }
//        for(int i = 0; i < n*k; i++)
//        	System.out.println(a1[i]);
        for(int i = 0; i < n*k-1; i++) {
            for(int j = 1 + i; j < n*k ; j++) {
            	count += a1[i] > a1[j] ? 1: 0;
            	//System.out.println(a1[i] + ", " + a1[j]);
            }
        }

//      for(int i = 0; i < n-1 ; i++) {
//    	for(int j = 1 + i; j < n; j++) {
//    		count += a[i] > a[j] ? 1: 0;
//    		//System.out.println(a[i] + ", " + a[j]);
//    	}
//    }

        System.out.println((count % (Math.pow(10, 9) + 7)));
	}

}