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.*;

class Main{
    public static void main(String[] args) {
        Scanner stdIn = new Scanner(System.in);
        int n = stdIn.nextInt();
        long k = stdIn.nextLong();
        int[] a = new int[n];
        for(int i = 0;i<n;i++) {
            a[i] = stdIn.nextInt();
        }
        long count = 0;
        int[] b = new int[n];
        for(int i = 0;i<a.length;i++) {
            for(int j = i+1;j<a.length;j++) {
                if(a[i] > a[j]) {
                    b[i]++;
                }
            }
        }

        for(int i = 0;i<a.length;i++) {
            count += sum(b[i],k) % (1000000000+7);
            count = count % (1000000000+7);
        }
        System.out.println((count)%(1000000000+7));
    }

public static long sum(int a,long k) {
    long goukei = 0;
    goukei = a * (k/2*(k+1));
    return goukei;
    }
}