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.*;
import java.lang.*;
import java.io.*;
import java.math.*;
 
import java.util.*;
import java.lang.*;
import java.io.*;

@SuppressWarnings("unchecked")
public class Main implements Runnable {

  static BufferedReader in;
  static PrintWriter out;
 
  public static void main(String[] args) {
      new Thread(null, new Main(), "whatever", 1<<29).start();
  }
 
  public void run() {
    in = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out, false);
 
    try
    {
      // in = new BufferedReader(new FileReader("class_treasurer.txt"));
      // out = new PrintWriter("output.txt");

      int t,x1,n,i,j,MOD=1000000007;
      long k;
      String str;
      String[] token;

      str=in.readLine().trim();
      token=str.split(" ");
      n=Integer.parseInt(token[0]);
      k=Long.parseLong(token[1]);

      str=in.readLine().trim();
      token=str.split(" ");

      int[] a=new int[n];
      int[] full=new int[n];
      int[] nextOnly=new int[n];

      for(i=0;i<n;i++) {
        a[i]=Integer.parseInt(token[i]);
      }

      for(i=0;i<n;i++) {
        for(j=i+1;j<n;j++) {
          if(a[i]>a[j]) {
            nextOnly[i]++;
          }
        }

        for(j=0;j<n;j++) {
          if(a[i]>a[j]) {
            full[i]++;
          }
        }
      }

      long total=0L;
      long nc2=((k-1L)*k)/2;
      nc2%=MOD;
      for(i=0;i<n;i++) {
        total+=((k*nextOnly[i])%MOD);
        total%=MOD;
        total+=((nc2*full[i])%MOD);
        total%=MOD;
      }

      out.println(total);

      out.flush();
      out.close();
    }
    catch(Exception e) {
      e.printStackTrace();
    }
  }
}