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.
import java.util.*;
import java.io.*;

public class Main{

    BufferedReader in;
    StringTokenizer str = null;
    PrintWriter out;

    private String next() throws Exception{
    	while (str == null || !str.hasMoreElements())
    	    str = new StringTokenizer(in.readLine());
    	return str.nextToken();
    }
    
    private int nextInt() throws Exception{
	return Integer.parseInt(next());
    }
    
    public void run() throws Exception{
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);
	int n = nextInt(), k = nextInt();
	if (n == k){
	    out.println(-1);
	    out.close();
	    return;
	}
	/*
	if (n == k + 1){
	    for(int i=1;i<=n;i++){
		out.print(i + " ");
	    }
	    out.println();
	    out.close();
	    return;
	}
	*/
	int []a = new int[n];
	for(int i=0;i<n;i++) a[i] = i+1;
	int good = n-1, j = n-1;
	while(good - k > 1){
	    int t = a[j];
	    a[j] = a[j-1];
	    a[j-1] = t;
	    good-=2;
	    j-=2;
	}
	if (good > k){
	    int t = a[0];
	    a[0] = a[j];
	    a[j] = t;
	}
	for(int i=0;i<n;i++){
	    out.print(a[i] + " ");
	}
	out.println();
        out.close();
    }

    public static void main(String args[]) throws Exception{
	new Main().run();
    }
}