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.*;
import static java.lang.Math.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

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

		// 計算
		long result = 0;
		long cnt = 0;
		long cnt2 = 0;
		ArrayList<Integer> list = new ArrayList<Integer>();
		for(int i = 0; i < n; i++){
		    for(int j : list){
		        if(j > a[i]){
		            cnt++;
		        }else if(j < a[i]){
		            cnt2++;
		        }
		    }
		    list.add(a[i]);
		}
		

        result = cnt*k + (cnt + cnt2)*k*(k-1)/2;
        
        //long k_ = k;
        //long k_1 = (k - 1);
        //long c_ = (cnt + cnt2);
        //long t = (c_*k_)%1000000007;
        //long result2 = cnt*k_ + t*k_1/2;

		// 出力
		//System.out.println(cnt);
		//System.out.println(cnt2);
		//System.out.println(result);
	    System.out.println(result%1000000007);
	    //System.out.println(floorMod(result, 1000000007L));
		//System.out.println(result2);
	    //System.out.println(result2%1000000007);
		//System.out.println(k_1);
		//System.out.println(t);
		//System.out.println();
	}

}