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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        // int n = Integer.parseInt(bf.readLine());
        StringTokenizer st = new StringTokenizer(bf.readLine());
        // int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());

        st = new StringTokenizer(bf.readLine());
        int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());

        int invcount = 0;
        for(int i=0; i<n; i++)  {
          for(int j=0; j<n-1; j++) {
            if(a[j] > a[j+1]) {
              int temp = a[j];
              a[j] = a[j+1];
              a[j+1] = temp;
              invcount++;
            }
          }
        }
        int val = n*n;
        int count = 1;
        for(int i=1; i<n; i++) {
          if(a[i] == a[i-1]) count++;
          else {
            val -= count*count;
            count=1;
          }
        }
        val -= count*count;
        val /= 2;
        int mod = 1000000007;
        long ans = 1L*k*invcount + (1L*k*(k-1)/2 % mod)* val;
        ans %= mod;
        out.println(ans);
        out.close(); System.exit(0);
    }
}