Problem: 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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;



public class Task{
    public static void main(String[] args) throws IOException{
        new Task().run();
    }
    
    StreamTokenizer in; 
    Scanner ins;
    PrintWriter out;

    int nextInt() throws IOException{
        in.nextToken();      
        return (int)in.nval;
    }
    
    long nextLong() throws IOException{
        in.nextToken();      
        return (long)in.nval;
    }
    
    double nextDouble() throws IOException{
        in.nextToken();      
        return in.nval;
    }
    
    char nextChar() throws IOException{
        in.nextToken();
        return (char)in.ttype;
    }
    
    String nextString() throws IOException{
        in.nextToken();
        return in.sval;
    }
     
    private static final String INPUT = "g.in";
    
    void run() throws IOException{        
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        ins = new Scanner(System.in);        
        out = new PrintWriter(System.out);        
        
        try{
            if(System.getProperty("xDx") != null){
                in = new StreamTokenizer(new BufferedReader(new FileReader("input.txt")));
                ins = new Scanner(new FileReader("input.txt"));
                out = new PrintWriter(new FileWriter("output.txt"));
            }else{
    //                in = new StreamTokenizer(new BufferedReader(new FileReader(INPUT)));
    //                ins = new Scanner(new FileReader(INPUT));
    //                out = new PrintWriter(System.out);
            }
        }catch(Exception e){
//            in = new StreamTokenizer(new BufferedReader(new FileReader(INPUT)));
//            ins = new Scanner(new FileReader(INPUT));
//            out = new PrintWriter(System.out);
        }
        int n = nextInt(), k = nextInt();
        if(k >= n){
            out.print(-1);
        }else{
            int pos = n - k;
            for(int i = 1; i <= n; i++){
                if(i < pos){
                    out.printf("%d ", i + 1);
                }else if(i > pos){
                    out.printf("%d ", i);
                }else{
                    out.printf("1 ");
                }
            }
        }
        out.close();
    }   
   
}