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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class r210_b {
	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
public static void main(String[] args)throws IOException {
	br=new BufferedReader(new InputStreamReader(System.in));
	pw=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); 
	int n=nextInt(),k=nextInt(),a[]=new int[n+1];
	if(n==k){
		pw.print(-1);
		pw.close();
		return;
	}
	for (int i =n-k+1; i <=n; i++) {
		a[i]=i;
	}
	a[n-k]=1;
	for (int i = 1; i <=n-k-1; i++) {
		a[i]=i+1;
	}
	for (int i = 1; i <=n; i++) {
		pw.print(a[i]+" ");
	}
	pw.close();
	}
private static int nextInt()throws IOException{
	return Integer.parseInt(next());
}
private static long nextLong()throws IOException{
	return Long.parseLong(next());
}
private static double nextDouble()throws IOException{
	return Double.parseDouble(next());
}
private static String next()throws IOException {
	while (st==null||!st.hasMoreTokens())
	 st=new StringTokenizer(br.readLine());
	return st.nextToken();
}
}
