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;

public class Main {

    public static class A {
        public int num;
        public int count = 0;
        public int beforeCount = 0;
        public A(){}
    }

    public static long siguma( long a ){
        long ans = 0 ;
        for( long i = 1 ; i <= a ; i++ ){
            ans += i;
        }
        return ans;
    }

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        String[] params = in.nextLine().split(" ");
        int N = Integer.parseInt(params[0]);
        long K = Long.parseLong(params[1]);
        params = in.nextLine().split(" ");

        A[] As = new A[N];
        for( int i = 0 ; i < N ; i++ ){
            As[i] = new A();
            As[i].num = Integer.parseInt(params[i]);
        }

        for( int i = 0 ; i < N ; i++ ){
            for( int j = 0 ; j < N ; j++ ){
                if( i == j ){
                    continue;
                }
                if(As[i].num > As[j].num){
                    As[i].count++;
                    if( j < i ){
                        As[i].beforeCount++;
                    }
                }
            }
        }

        long count = 0;
        long s = siguma(K);
        for( int i = 0 ; i < N ; i++ ){
            count += As[i].count * s;
            while( count / 1000000007 > 0 ){
                count -= 1000000007;
            }
            count -= As[i].beforeCount * K;
            while( count / 1000000007 < 0 ){
                count += 1000000007;
            }
        }

        System.out.println(count%1000000007);

    }

}