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.
/**
 * Author   : KillerCage
 * Language : Java
 * Time     : 2014 / 08
 * Quote    : E'la nostra ora incisa sull'anello
 */

import java.util.*;
import java.io.*;
import java.math.*;

public class Main {

	static InputStreamReader inp = new InputStreamReader( System.in );
	static BufferedReader buf = new BufferedReader( inp );
	static StreamTokenizer tok = new StreamTokenizer( buf );

	static OutputStreamWriter out = new OutputStreamWriter( System.out );
	static PrintWriter wri = new PrintWriter( out );

	static int nextInt() throws IOException {
		tok.nextToken(); return ( int ) tok.nval;
	}

	public static void main( String[] args ) throws IOException {
		int n = nextInt();
		int k = nextInt();
		if ( k >= n ) {
			wri.println( "-1\n" );
		} else {
			for ( int i = 1; i < n - k; i ++ ) {
				wri.print( ( i + 1 ) + " " );
			}
			wri.print( 1 + " " );
			for ( int i = n - k + 1; i <= n; i ++ ) {
				wri.print( i + " " );
			}
			wri.println( "" );
		}
		wri.flush();
	}
}
