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.*;
public class fence {
 //  static{System.out.println("hello"); }
   public static void main(String args[])
   { Scanner scn=new Scanner(System.in);
 int n=scn.nextInt(); 
 int k=scn.nextInt(); int a[]=new int[n+1];
 Arrays.fill(a,-1);
 if (k>=n) {System.out.println(-1); System.exit(0);}
 int i;
 for(i=2 ; i<=k+1 ; i++) a[i]=i; a[1]=1;
 if (k!=n-1){
int temp=i;
for(;i<n;i++) a[i]=i+1;
a[1]=temp; a[n]=1;}
 for( i=1;i<=n;i++) System.out.print(a[i]+" ");
}
}
