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.io.*;
import java.math.*;
import java.util.*;
public class Main { 
 
	public static void main(String[] args) { 
		Scanner input = new Scanner(System.in);
		long mod = (long)(1e9+7);
		int N = input.nextInt();
		long K = input.nextLong();
		long ans = 0;
		long choose2Right = K*(K+1)/2; //Sum of first N numbers is N*(N+1)/2;
		//For K concatenations, sum of 1 through K pairs exist
		K--;
		long choose2Left = K*(K+1)/2;
		choose2Right%=mod;
		choose2Left%=mod;
		int[] arr = new int[N];
		for (int i = 0; i < N; i++) {
			arr[i] = input.nextInt();
		}
		for (int cur = 0; cur < N; cur++) {
			int curNum = arr[cur];
			long left = 0;
			long right = 0;
			for (int L = 0; L < cur; L++) {
				if (curNum>arr[L]) left++;
			}
			for (int R = cur+1; R < N; R++) {
				if (curNum>arr[R]) right++;
			}
			long curAns = right*choose2Right;
			curAns%=mod;
			curAns+=left*choose2Left;
			curAns%=mod;
			ans+=curAns;
			ans%=mod;
		}
		System.out.println(ans);
	}
}