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) {

		long ans = 0;

		Scanner sc = new Scanner(System.in);
		int  n = sc.nextInt();
		long  k = sc.nextLong();
//		longd = sc.nextInt();
//
//		boolean f = false;
		int[] ai = new int[n];
//		int[] tmp = new int[n];
		for (int i = 0; i < n; i++) {
			ai[i] = sc.nextInt();
//			tmp[i] = 0;
		}

//		int[] bi = new int[n];
//		for (int i = 0; i < k; i++) {
//			bi[i] = sc.nextInt();
//		}

		long tmp = 0;
		long tmp2 = 0;

		for(int i=0; i < n; i++) {
			for(int j = i+1; j < n; j++){
				if(ai[i] > ai[j]) {
					tmp++;
				}
				if(ai[i] < ai[j]) {
					tmp2++;
				}
			}
		}
//		System.out.println(tmp);

		for(long i = 1; i <= k; i++) {
			if(i == 1) {
				ans = (i * tmp) + ans;
			} else {
				ans = (i * tmp) + ans + ((i-1) * tmp2);
			}
		}
//		System.out.println(ans);
		ans = ans % 1000000007;

		System.out.println(ans);

		sc.close();
	}
}