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.Arrays;
import java.util.Scanner;


public class Permutation {

    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        
        int n = input.nextInt(), kAns = input.nextInt(), k = n - 1;
        
        if(n == kAns){
            System.out.println(-1);
            System.exit(0);
        }
        
        int[] res = new int[n];
        for(int a=0; a<res.length; a++)
            res[a] = a+1;
        
        if(kAns % 2 != k % 2){
            swap(res, 0, n-1);
            k--;
        }
        
        int at = 1;
        while(k > kAns){
//            System.out.println(Arrays.toString(res)+" "+at);
            swap(res, at);
            do{
                at++;
            }while(at < res.length && res[at] != at+1);
            k-=2;
        }
        
        for(int i : res)
            System.out.print(i+" ");
        
        input.close();
    }
    
    public static void swap(int[] arr, int i1, int i2){
        int temp = arr[i1];
        arr[i1] = arr[i2];
        arr[i2] = temp;
    }
    
    // first and swap are indexes
    public static void swap(int[] arr, int first){
        int swap = first + 1;
        while(true){
            if(arr[swap] == swap + 1 && gdc(arr[swap], arr[first]) == 1){
                swap(arr, first, swap);
                return;
            }
            swap++;
        }
    }
    
    public static int gdc(int a, int b){
        if(a > b){
            int temp = a;
            a = b;
            b = temp;
        }
        if(a == 0)
            return b;
        return gdc(b%a, a);
    }
    
}
