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 mod = (long) (Math.pow(10.0, 9.0) + 7);
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long k = in.nextInt();
		int a[] = new int[n];

		for ( int i = 0; i < a.length; i++ ) {
			a[i] = in.nextInt();
		}

		long inner = 0; // 内部で成立する転倒数の個数
		long outer = 0; // 他のｎとの関係で成立しうる転倒数の個数
		for ( int i1 = 0; i1 < n; i1++ ) {
			for ( int i2 = 0; i2 < n; i2++ ) {
				if ( a[i1] > a[i2] ) {
					outer++;
					if ( i1 < i2 ) {
						inner++;
					}
				}
			}
		}
		long kk = (k - 1) * k / 2 % mod;
		long ans = (inner * k) % mod + (outer * kk) % mod;
		System.out.println(ans);
		in.close();
	}
}