Problem: Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class LevkoandPermutation
{

    
    /************************ SOLUTION STARTS HERE ***********************/

    private static void swap(int arr[] , int i , int j){
	int temp = arr[i];
	arr[i] = arr[j];
	arr[j] = temp;
    }
    
    private static void solve(FastScanner s1, PrintWriter out){

	int n = s1.nextInt();
	int k = s1.nextInt();
	if(k == n)
	    out.print(-1);
	else{
	    int max = n - k;
	    int arr[] = new int [n];
	    for(int i=0;i<n;i++)
		arr[i] = i+1;

	    for(int i = (max % 2 == 0) ? 0 : 1;i<n-k;i += 2)
		   swap(arr, i, i+1);
		
	    for(int a:arr)
		out.print(a + " ");
	}
    }


    
    /************************ SOLUTION ENDS HERE ************************/
    
    
    
    
    
    /************************ TEMPLATE STARTS HERE *********************/

    public static void main(String []args) throws IOException {
	FastScanner in  = new FastScanner(System.in);
	PrintWriter out = 
		new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false); 
	solve(in, out);
	in.close();
	out.close();
    }    

    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}	
	String next()
	    {while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}		    
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}	    	  	
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}		
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}	
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}	    	
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}	    	
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}				
    }

    /************************ TEMPLATE ENDS HERE ************************/
}