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.io.PrintWriter;
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        new Solver().solve(in, out);

        in.close();
        out.flush();
    }

    private static class Solver {
        public void solve(Scanner in, PrintWriter out) {
            int length = in.nextInt();
            int re = in.nextInt();
            int[] ints = new int[length];

            for (int i = 0; i < length; i++) {
                ints[i] = in.nextInt();
            }

            int a = 0;
            int b = 0;
            for (int i = 0; i < length; i++) {
                for (int j = 0; j < length; j++) {
                    if (ints[i] <= ints[j]) continue;
                    a++;
                    if (j > i) b++;
                }
            }
            int mod = 1000000007;
            long counter = 0;
            counter += ((long) b * re) % mod;
            counter += (a * 0.5 * ((re * (re - 1L)) % mod)) % mod;

            out.println(counter % mod);
        }
    }
}