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: Time Limit Exceeded)

import java.util.*;
import java.util.stream.Stream;

public class Main {

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);

		String[] lines = scanner.nextLine().split(" ");
		int N = Integer.parseInt(lines[0]);
		int K = Integer.parseInt(lines[1]);

		String[] arrays = scanner.nextLine().split(" ");
		int[] newArray = Stream.of(arrays).mapToInt(Integer::parseInt).toArray();
		ArrayList<Integer> array = new ArrayList<Integer>();
		for (int i = 0; i < newArray.length; i++)
			array.add(newArray[i]);

		int length = array.size();

		for (int i = 0; i < K - 1; i++) {
			for (int j = 0; j < length; j++) {
				array.add(array.get(j));
			}
		}
		
		
		int B = 0;
		for (int i = 0; i < array.size(); i++) {
			for (int j = i + 1; j < array.size(); j++) {
				if (array.get(i) > array.get(j))
					B++;
			}
		}

		System.out.println(B % 1000000007);

	}

}