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.IOException;
import java.io.InputStreamReader;

public class Main {


	public static void main(String[] args) throws IOException {
		BufferedReader bf = null;
		bf = new BufferedReader(new InputStreamReader(System.in));
		int n;
		long k,temp = 0,b = 0,ans = 0,num;
		int[] a;
		String str;
		String[] s;
		str = bf.readLine();
		s = str.split(" ");
		n = Integer.parseInt(s[0]);
		k = Long.parseLong(s[1]);
		str = bf.readLine();
		s = str.split(" ");
		a = new int[n];
		for(int i = 0;i < n;++i) {
			a[i] = Integer.parseInt(s[i]);
			for(int j = 0;j < i;++j) {
				if(a[j] > a[i]) {
					b++;
					temp++;
				} else if(a[j] < a[i]) {
					temp++;
				}
			}
		}
		ans = temp;
		if(k%2 == 0) {
			num = k/2;
			num *= k-1;
		} else {
			num = (k-1)/2;
			num *= k;
		}
		num %= 1e9+7;
		ans *= num;
		ans %= 1e9+7;
		num = b*k;
		num %= 1e9+7;
		ans += num;
		ans %= 1e9+7;
		System.out.println(ans);
	}
}