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


@SuppressWarnings("unused")
public class B {
	public  static  FastScannerB scan = new FastScannerB();
	public static PrintWriter out = new PrintWriter(System.out);
	
	public static void solve () {
		int n=scan.nextInt(),k=scan.nextInt();
		if(n==k) {
			out.println(-1);
		}
		else {
			out.print(n-k+" ");
			for(int i=1;i<n-k;i++) {
				out.print(i+" ");
			}
			for(int i=n-k+1;i<=n;i++) {
				out.print(i+" ");
			}
		}
		
		
	}
	public static void main(String[] args) {
		
		solve();
		out.close();
	}

}
class FastScannerB {
	BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st= new StringTokenizer("");
	String next() {
		while (!st.hasMoreTokens())
			try {
				st=new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		return st.nextToken();
	}
	int nextInt() {
		return Integer.parseInt(next());
	}
	int[] readArray(int n) {
		int[] a=new int[n];
		for (int i=0; i<n; i++) a[i]=nextInt();
		return a;
	}
	long nextLong() {
		return Long.parseLong(next());
	}
	void sort(long[] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	

	
}


