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.*;
public class LevkoAndPermutation {
    public static void main(String args[]){
        Scanner sc =new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        if(k==n)
            System.out.println(-1);
        else{
            ArrayList<Integer>al=new ArrayList<>();
            if(k==n-1){
                for(int i=0;i<n;i++)
                    al.add(i+1);
            }
            else{
                int s=2;
                for(int i=0;i<n;i++){
                    al.add(s);
                    if(s==n-k)
                        break;
                    s++;
                }
                al.add(1);
                s=n-k+1;
                while(al.size()<n){
                    al.add(s);
                    s++;
                }
            }
            for(int i=0;i<al.size();i++)
                System.out.print(al.get(i)+" ");
        }
    }
}
