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.IOException;
import java.io.InputStreamReader;


public class prem{
	public static void main(String[]args) throws IOException{
	BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	String x = bf.readLine(); 
	boolean can =true;
	String[] xxi = x.split(" ");
	long n = Long.parseLong(xxi[0]);
	long k = Long.parseLong(xxi[1]);
	long [] xx = new long[(int) n];
	
	int counter =0; 
	for(int i =1; i<n; i++){
		if(counter<k){
			xx[i]=(i+1);
		}
		else{
			if(counter==k){
				xx[i]=1;
			}
			else{
				xx[i]=(i+1)-1; 
			}
			
		}
		counter++;
	}
	if(n-k==1){
		xx[0]=1;
	}
	else{
	xx[0]=n; 
	
	}
	if(n==1 && k==0){
		System.out.println(1);
	}
	else{
	if(n==1|| k>n || n==k ){
		System.out.println(-1);
	}
	else{
	for(int i =0; i<n; i++){
		System.out.print(xx[i]+ " ");
	}
	}
	}}
}

