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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {  
    
    public static void swap(int []arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;            
    }
   
    public static void main(String [] args) throws IOException {
        
        BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();        
        String [] ar = input.readLine().split(" ");
        int n = Integer.parseInt(ar[0]);
        int k = Integer.parseInt(ar[1]);
        int arr[] = new int[n];
        
        if(k >= n) {
            System.out.println(-1);
            return;
        }
        
        for(int i = 1;i<=n;i++){
            arr[i-1] = i;
        }
        if((n-k-1) > 0) {
            int rem = n-k-1;
            if(rem % 2 == 0) {
                for(int i=k+1;i<arr.length-1;i+=2) {
                    swap(arr, i, i+1);
                }
            } else {    
                swap(arr, k+1,0);
                for(int i=k+2;i<arr.length-1;i+=2) {
                    swap(arr, i, i+1);
                }                
            }
        }
        for(int v: arr) {
            output.append(v);
            output.append(" ");
        }
        System.out.print(output);
       
        
        
        
        
        
        
    }
       
}
    