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;
import java.util.Arrays;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		long k=sc.nextLong();
		int[] a=new int[n];
		long mod=1000000007;
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
		}
		long count=0;
		for(int i=0; i<n; i++){
			int aa=a[i];
			for(int j=i+1; j<n; j++){
				if(aa>a[j]){
					count++;
				}
			}
		}
		Arrays.sort(a);
		int min=a[0];
		int pp=0;
		long uu=0;
		for(int i=1; i<n; i++){
			if(min==a[i]){
				uu+=pp;
			}else{
				min=a[i];
				pp++;
				uu+=pp;
			}
		}
		System.out.println(uu);
		long hhh=(((k-1)*k)/2)%mod;
		System.out.println(((k*count)%mod+((hhh)*uu)%mod)%mod);
	}
}