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.*;
import java.io.*;


public class lvp
{   

   public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
  int n=sc.nextInt();
  int k=sc.nextInt();
  if(n==k){
     System.out.println(-1);
  }else if(k==n-1){
    for(int i=1;i<=n;i++){
        System.out.print(i+" ");
    }
    }else{
      int arr[]=new int[n+1];
      for(int i=2;i<=2+k-1;i++){
        arr[i]=i;
      }
      arr[n]=1;
      for(int i=n-1;i>=0;i--){
        if(arr[i]!=0){
            arr[1]=i+1;
            break;
        }else{
            arr[i]=i+1;
        }
      }
      for(int i=1;i<=n;i++){
        System.out.print(arr[i]+" ");
      }
    }
  }

}
