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.*;

public class Main {
	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        StringBuilder sb = new StringBuilder();
        sc.nextLine();
        String[] array = sc.nextLine().split(" ");
        int length = array.length;
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < length; j++) {
                sb.append(array[j]);
            }
        }
        int size = sb.length();
        int cnt = 0;
        for (int i = 0; i < size; i++) {
            int tmp = Character.getNumericValue(sb.charAt(i));
            for (int j = 0; j < size; j ++) {
                if (i == j || i > j) continue;
                if (tmp > Character.getNumericValue(sb.charAt(j))) {
                    cnt++;
                }
            }
        }
        System.out.println(cnt % 1000000007);
    }
}