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.*;

class Main{

    public static void main(String[] args){

	Scanner sc = new Scanner(System.in);

	int N = sc.nextInt();

	int K = sc.nextInt();

	int[] A = new int[N];

	int preans = 0;

	int ans = 0;

	for(int i=0; i<N; i++){

	A[i] = sc.nextInt();

  }

	for(int i=1; i<N; i++){

	for(int j=0; j<i; j++){

	if(A[j]>A[i]){

	preans++;

    }

   }

   }

	for(int i=0; i<N; i++){

	for(int j=0; j<N; j++){

	if(A[i]>A[j]){

	ans++;

    }

   }

   }		

	int B = 0;	

	for(int i=1; i<K; i++){

	B += i;

  }



	int lastans = preans*K + B*ans;

	System.out.println(lastans%(1000000000+7));

 }
}