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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
            Scanner sc = new Scanner(System.in);
            
            long BIG = 10;
            for(int i = 0;i<8;i++){
                BIG*=10;
            }
            BIG+=7;
            
        
            int N = sc.nextInt();
            long K = sc.nextLong();
            
            int[] arr = new int[N];
            
            for(int i = 0; i < N; i++){
                arr[i] = Integer.parseInt(sc.next());
            }
            
            long count1 = 0;
            long count2 = 0;
            
            // 次以降
            for(int i = 0; i < N; i++){
                for(int j = 0; j < N; j++){
                    if(arr[i] > arr[j]){
                        count1++;
                        count1 %= BIG;
                    }
                }   
            }
            count1 %= BIG;
            long result = add(K) % BIG;
            count1 *= result;
            
            // 自分のなか
            for(int i = 0; i < N-1; i++){
                for(int j = i+1; j < N; j++){
                    if(arr[i] > arr[j]){
                        count2++;
                    }
                }   
            }
            count2 %= BIG;
            K %= BIG;
            count2*=K;
            System.out.println((count1+count2)%BIG);
    }
    
    static long add(long K){
        long sum = 0;
        for(int i = 1; i < K; i++){
            sum += i;
        }
        return sum;
}
}