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 java.util.ArrayList;

public class Main{
	public static void main(String args[]){

		// 宣言
		int mod = 1_000_000_007;
		ArrayList<Integer> A = new ArrayList<>();
		ArrayList<Integer> B = new ArrayList<>();
		Scanner sc = new Scanner(System.in);
		int num = 0;
		// 値取得
		int N = sc.nextInt();
		long K = sc.nextInt();
		int Ble = (int)K * N;
		// System.out.println(N);
		for(int i=0;i<N;i++){
			A.add(sc.nextInt());
			// System.out.println(A.get(i));
		}

		// リストB作成
		int index = 0;
		int cnt=0;
		while(cnt<Ble){

			B.add(A.get(index));
			if(index == N - 1){
				index = 0;
			}else{
				index++;
			}
			// System.out.println(B.get(cnt));
			cnt++;
		}
		// 転倒数を求める(自分より右にある小さい数字を足していく)
		int ten = 0;
		long ans = 0L;
		for(int k = 0;k<Ble;k++){
			for(int l = 0;l<Ble;l++){
				if(k < l){
					if(B.get(k) > B.get(l)){
						ten = ten + B.get(l);
					}					
				}				
			}
		}
		System.out.println(ten % mod);
        sc.close();
	}
}