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.*;

public class Main{
  public static void main(String[]args){
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    long K = sc.nextLong();
    int mod = 1000000007;
    int[] num = new int[N];
    for(int i = 0; i < N; i++){
      num[i] = Integer.parseInt(sc.next());
    }
    
    long ans = 0;
    for(int i = 0; i < N; i++){
      long n1 = 0;
      long n2 = 0;
      for(int j = 0; j < N; j++){
        if(num[i] > num[j]){
          if(i < j){
            n1++;
          }else if(i > j){
            n2++;
          }
        }
      }
      
      ans += (n1 * ((K*(K+1)/2)%mod)) % mod;
      ans += (n2 * ((K*(K-1)/2)%mod)) % mod;
      ans %= mod;
    }
    System.out.println(ans);
  }
}