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.
/**
 * Created by Omar on 2/9/2016.
 */

import java.util.*;
import java.io.*;

public class Levko {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        String[] parts=br.readLine().split(" ");

        int n=Integer.parseInt(parts[0]);
        int k=Integer.parseInt(parts[1]);

      if(n==k){
          pw.print("-1\n");
      }
        else{
          for(int i=2;i<=n-k;i++){
              pw.print(i);
              pw.print(" ");
          }
          pw.print("1 ");
          for(int i=n-k+1;i<=n;i++){
              pw.print(i);
              pw.print(" ");
          }
      }


        pw.close();
        br.close();
    }
    public  static int GCD(int a, int b) { return b==0 ? a : GCD(b, a%b); }
}
