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.
//package Contest_Div210;
//Xudo xoxlasa Accepted
import java.util.Scanner;

public class B {
public static void main(String[] args) {
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int k=sc.nextInt();
	if(n==k){
		System.out.println("-1");
		return;
	}
	int [] a=new int[n+1];
	int y=1;
	for (int i =1; i <n-k; i++) {
		a[i]=i+1;
	}
	a[n-k]=1;
	for (int i =n-k+1; i <=n; i++) {
		a[i]=i;
	}
	for (int i =1; i <=n; i++) {
		System.out.print(a[i]+" ");
	}
	}
}
