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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        B solver = new B();
        solver.solve(1, in, out);
        out.close();
    }

    static class B {
        static final long mod = 1000000007L;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            int n = Integer.parseInt(in.next());
            long k = Long.parseLong(in.next());
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(in.next());
            }
            long ans = 0;
            for (int i = 0; i < n; i++) {
                long right = 0;
                long all = 0;
                for (int j = i; j < n; j++) {
                    if (a[i] > a[j]) right++;
                }
                for (int j = 0; j < n; j++) {
                    if (a[i] > a[j]) all++;
                }
                ans += right * k % mod;
                ans %= mod;
                ans += (k - 1) * k / 2 % mod * (all % mod);
                ans %= mod;

            }

            // 出力
            out.println(ans);

        }

    }
}