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.Scanner;

public class Main {
	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt();
			int K = sc.nextInt();
			int[] nums = new int[N];
			for (int i = 0; i < N; i++) {
				nums[i] = sc.nextInt();
			}
			long[] childCount = new long[N];
			for (int i = N - 1; i > 0; i--) {
				for (int j = i - 1; j >= 0; j--) {
					if (nums[j] > nums[i]) {
						childCount[j] = (childCount[j] + childCount[i]) + 1;
					}
				}
			}
			long div = 1000000007;
			long k;
			if( K %2 == 1) {
				long amari1 = K % div;
				long amari2 = (K+1)/2 % div;
				k = amari1 * amari2 % div;
			}else {
				long amari1 = K/2 % div;
				long amari2 = (K+1) % div;
				k = amari1 * amari2 % div;
			}
			System.out.println(childCount[0] * k % div);
		}
	}
}