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.Scanner;

public class Main {
    public static void main(String[] args) {
        new Main().run();
    }

    private void run() {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        long K = scanner.nextLong();
        int[] A = new int[N];

        long g = 0;
        long h = 0;
        long tmp = 0;
        long tmp2 = 0;
        for (int i = 0; i < N; i++) {
            int f = scanner.nextInt();
            A[i] = f;
        }
        for (int j = 0; j < N - 1; j++) {
            for (int i = j; i < N; i++) {
                if(A[j] > A[i]) {
                    g++;
                }
            }
        }
        for (int j = 0; j < N; j++) {
            for (int i = 0; i < N; i++) {
                if(A[j] > A[i]) {
                    h++;
                }
            }
        }

        g += tmp;
        h += tmp2;

        long ans = ((g % 1000000007) * (K % 1000000007)) % 1000000007;
        ans = (ans + (h % 1000000007) * ((K*(K-1)/2) % 1000000007)) % 1000000007;
//        ans = (ans + (total % 1000000007) * (m % 1000000007)) % 1000000007;

        System.out.println(ans);

    }
}