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.util.*;
public class Main{
    public static void main(String[] args) throws IOException{
        MyReader r = new MyReader();  
        int N;
        long K; 
        {int[] a = r.ii(); N = a[0]; K = a[1];}
        int[] A = r.ii();
        int MOD = 1000000007;
        int n1 = 0;
        int n2 = 0;
        for (int i = 0; i < N; i++) {
            long Ai = A[i];
            for (int j = 0; j < i; j++)
                if(Ai > A[j])
                    n1++;
            for (int j = i+1; j < N; j++)
                if(Ai > A[j])
                    n2++;
        }
        long num = K*(K-1)/2%MOD*n1+K*(K+1)/2%MOD*n2;
        print(num%MOD);
    }
    static void print(Object o){
        System.out.print(o.toString());
    }
    static void println(Object o){
        System.out.println(o.toString());
    }
    static int Int(String s){
        return Integer.parseInt(s);
    }
    static long Long(String s){
        return Long.parseLong(s);
    }
    static class MyReader extends BufferedReader{
        MyReader(){
            super(new InputStreamReader(System.in));
        }
        String s() throws IOException{
            return readLine();
        }
        String[] ss() throws IOException{
            return s().split(" ");
        }
        int i() throws IOException{
            return Int(s());
        }
        int[] ii() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            int[] ii = new int[size];
            for(int j = 0; j < size; j++) ii[j] = Integer.parseInt(ss[j]);
            return ii;
        }
        long l() throws IOException{
            return Long(s());
        }
        long[] ll() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            long[] ll = new long[size];
            for(int j = 0; j < size; j++) ll[j] = Long.parseLong(ss[j]);            
            return ll;
        }
    }
}