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.Scanner;
//import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
class Main {
    public static void main(String[] args) {
        Scanner stdIn = new Scanner(System.in);
        
        int n  = stdIn.nextInt();
        int  k = stdIn.nextInt();

        int[] a = new int[n];
        for(int i = 0;i < n;i++) {
            a[i] = stdIn.nextInt();
        }

        long count= 0;
        long t = 1000000007;

        long x = 0;
        long y = 0;
        for(int i = 0;i < n;i++) {
            for(int j = i+1;j < n;j++){
                if(a[i] > a[j]){
                    x++;
                }
            }
        }
        for(int i = 0;i < n;i++) {
            for(int j = i+1;j < n;j++){
                if(a[i] != a[j]){
                    y++;
                }
            }
        }

        count = (((x * k) % t) + ((((y * k) % t) * (k-1) / 2) % t)) % t;

        
        System.out.println(count);

        
    }

}