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.Scanner;


public class B {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int t [] = new int[n+1];
		for(int i = 1; i <= n; i++) t[i] = i;
		int shifts = n-k;
		for(int i = 1; i <= shifts; i++){
			t[i] = i+1;
		}
		t[shifts] = 1;
		int count = 0;
		for(int i = 1; i <=n; i++){
			if(gcd(i, t[i]) > 1) count ++;
		}
		if(count == k){
			for(int i = 1; i <=n; i++){
				System.out.print(t[i] + " ");
			}
		} else{
			System.out.println("-1");
		}
		
	}
	
	public static int gcd(int a, int b){
		if(a < b){
			int temp = a;
			a = b;
			b = temp;
		}
		if(b == 0) return a;
		return gcd(b, a % b);
	}

}
