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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = in.nextInt();
		int k = in.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = in.nextInt();
		System.out.println(solve(n, k, a));
		in.close();
	}

	private static long solve(int n, int kk, int[] a) {
		long ret = 0;
		long k = kk;
		for (int i = 0; i < n; i++) {
			long r1 = 0;
			long r2 = 0;
			for (int j = 0; j < i; j++)
				if (a[i] < a[j])
					r1++;
			for (int j = i + 1; j < n; j++)
				if (a[i] < a[j])
					r2++;
			ret += r1 * k;
			ret += (r1 + r2) * (k - 1) * k / 2;
			ret %= 1000000007;
		}
		return ret;
	}
}