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: Wrong Answer)

import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.rmi.server.UnicastRemoteObject;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main //implements Runnable
{
    static long mod=1000000007;
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};

    public static void main(String[] args)   throws Exception, IOException{        
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n=sc.nextInt();
        int k=sc.nextInt();
        int a[]=sc.nextIntArray(n);
        long d[]=new long[n];
        long e[]=new long[n];
        for (int i = 0; i < n; i++) {
        	for (int j = 0; j < a.length; j++) {
				if(a[i]<a[j]) {
					if(i>j)d[i]++;
					e[i]++;
				}
			}
		}
        // 2 1 2 1
        long s=0;
        for (int i = 0; i < d.length; i++) {
			s+=e[i]*((long)k*(k-1)/2%mod)%mod + d[i]*k%mod;
			s%=mod;
			if(s<mod)s+=mod;
//			db(i,s);
		}
        
        out.println(s);
     	out.flush();
    }
    
    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}

//class P {
//	int x, y;
//	P(int  x, int y) {
//	  this.x=x;
//	  this.y=y;
//	}
//}

//class P implements Comparable<P>{
//    int a,b;
//    P(int a, int b) {
//    	this.b=b;
//    	this.a=a;
//    }
//
//    public int compareTo(P p){
//    	if(a==p.a) return b-p.b;
//        return -p.a+a;
//    }
//}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) throws IOException{
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) throws IOException{
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}