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: Runtime Error)

import java.util.*;
public class Main{
    public static void main(String [] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        long[] A = new long[N];
        long[] B = new long[N*K];
        long count = 0;
        for(int i = 0; i < N; i++){
            A[i] = sc.nextLong();
        }
        for(int i = 0; i < K; i++){
            for(int j = 0; j < N; j++){
                B[j+i*N] = A[j];
            }
        }
for(int i = 0; i < N*K -1; i++){
for(int j = i+1; j < N*K; j++){
if(B[i] > B[j])count = (count+1)%1000000007;
}
}
System.out.println(count);
    }
}