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.util.*;
public class SinIsh {
public static void main(String [] args){
Scanner in = new Scanner(System.in);
int n=in.nextInt();
int k=in.nextInt();
int array[]=new int[n];
if(n==k)
System.out.println("-1");
if(n==k+1){
for(int i=0;i<n;i++){
System.out.print((i+1) + " ");
}
}
else if(n>k+1){
for(int i=0;i<n;i++){
if(i<=k)
array[i]=i+1;
if(i>k)
array[i]=i+2;
}
array[0]=k+2;
array[n-1]=1;
for(int i=0;i<n;i++){
System.out.print(array[i] + " ");
}
}
}
}