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.*;
import java.lang.*;
import java.io.*;

	public class FastIO{
			
		BufferedReader br;
		StringTokenizer st;
		
		public FastIO(){ // constructor
			
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next(){
			
			while (st == null || !st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				}
				
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt(){
			
			return Integer.parseInt(next());
		}
		
		long nextLong(){
			
			return Long.parseLong(next());
		}
		
		double nextDouble(){
			return Double.parseDouble(next());
		}

		String nextLine(){
			
			String str = "";
			try{
				
				str = br.readLine();
			}
			catch (IOException e){
				
				e.printStackTrace();
			}
			return str;
		}
	
	public static void main(String[] args){
		
		FastIO io = new FastIO();
		
		// int n = io.nextInt(); -> for int
		// double d = io.nextDouble(); -> for double
		// String ch = io.next(); -> for one word
		// String str = io.nextLine(); -> for the entire line
		
		int n = io.nextInt();
		int k = io.nextInt();
		
		if((n - k) == 0) {
			System.out.println(-1);
			return;
		}
		
		else {
			
			int a[] = new int[n + 1];
			
			for(int i = 1; i < n + 1; i++) {
				a[i] = i;
			}
			
			int gcd1 = n - k - 1;
				
			int i = 2;
			while(gcd1 > 0) {
				int temp = a[i];
				a[i] = a[i-1];
				a[i-1] = temp;
				i++;
				gcd1--;	
			}
			for(int idx = 1; idx < n + 1; idx++) {
				System.out.print(a[idx] + " ");
			}
		}
		
	}
}