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.*;

/**
 * Problem statement:
 * https://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=2733
 *
 * The online judge has extremely weak test data, so this solution is accepted.
 * However, there's a bug! If the test data were not ridiculously weak, this
 * code would get wrong answer. Can you find the problem?
 */
public class FindTheBug5 {

    public static void main(String[] args){
    	Scanner sc = new Scanner(System.in);
    	int n = sc.nextInt();
    	int k = sc.nextInt();
    	if(k >= n ){
    		System.out.println(-1);
    	}else{
    		// many are possible, just think of the easiest case
    		int[] data = new int[n+1];
    		if(k == n -1){
    			for(int i = 1; i <= n; i++){
    				System.out.println(i+" ");
    			}
    		}else{
    			data[1] = n;
    			//Get ks we want
    			for(int i = 2; i < k+2; i++){
    				data[i] = i;
    			}
    			data[k+2] = 1;
    			for(int i = k +3; i <=n; i++){
    				data[i] = i - 1;
    			}
    			for(int i = 1; i <= n; i++){
    				System.out.print(data[i] + " ");
    			}
    		}
    		
    	}
    	sc.close();
    }

}