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:
//package codeforces;

import java.io.IOException;
import java.util.Scanner;

public class cf210b
{
    public static void main(String args[]) throws IOException
    {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();
        int arr[] = new int[n];
        for(int i=0;i<n;i++)
        {
            arr[i] = i+1;
        }
        if(n==1 && k==1)
            System.out.println(-1);
        else if(k==n)
            System.out.println(-1);
        else
        {
            for(int i=0;i<(n-k-1);i++)
            {
                int temp = arr[i];
                arr[i] = arr[i+1];
                arr[i+1] = temp;
            }
            
            for(int i=0;i<n;i++)
            {
                System.out.print(arr[i]+" ");
            }
        }
        
    }
}