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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.SortedSet;
import java.util.TreeSet;

public class A {	

	static Scanner sc = new Scanner(System.in);
	static InputStreamReader stream = new InputStreamReader(System.in);
	static BufferedReader br = new BufferedReader(stream);
	
	public static void main(String[] args){	
		int n = sc.nextInt();
		int k = sc.nextInt();
		if(n==k) System.out.println(-1);
		else if(n-1==k) {
			int[] a = give(n);
			for(int e:a) {
				System.out.print(e + " ");
			}
			System.out.println();
		}
		else {
			int[] list = new int[n-k];
			int[] a = give(n);
			list[0]=1;
			int pointer = 1;
			for(int i=k+1 ; i<n ; i++) {
				list[pointer] = a[i];
				pointer++;
			}
			int p = k+1;
			for(int i=0 ; i<n-k ; i++) {
				a[p%n] = list[p-k-1];
				p++;
			}
			for(int e:a) {
				System.out.print(e + " ");
			}
			System.out.println();			
		}
				
	}
	
	public static int[] give(int n){
		int[] a = new int[n];
		for(int i=1 ; i<=n ; i++) {
			a[i-1] = i;
		}
		return a;
	}
	
	public static int pow(int n, int k) {
		int pow = 0;
		while(n%k==0 && n!=1) {
			n/=k;
			pow++;
		}
		if(n==1) return pow;
		else return -1;
	}
	
	//BufferedReader Input functions
		public static int readInt() throws IOException{
			return (Integer.parseInt(br.readLine()));
		}
		
		public static int[] readArray(int n) throws IOException{
			int[] a = new int[n];
			for(int i=0 ; i<n ; i++) {
				a[i]= readInt();
			}
			return a;
		}
		
		public static int[] readArrayline(int n) throws IOException{
			int[] a = new int[n];
			String[] s = br.readLine().split(" ");
			for(int i=0 ; i<n ; i++) {
				a[i]= Integer.parseInt(s[i]);
			}
			return a;
		}
		public static int[] readArraylinesc(int n){
			int[] a = new int[n];
			for(int i=0 ; i<n ; i++) {
				a[i]= sc.nextInt();
			}
			return a;
		}
	


}
