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.util.regex.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.lang.Integer.*;
import static java.lang.Double.*;
import static java.util.Collections.*;
import java.io.*;

public class _361_B_Levko_and_Permutation {
	//->solution screencast http://youtu.be/oHg5SJYRHA0
	public void solve() {
		int n = ni(), k=ni();
		if(n==k){
			out.print(-1);
			return;
		}
			
		int [] a = new int[n];
		for (int i = 0; i < a.length; i++) {
			a[i]=i+1;
		}
		for (int i = k+1; i <= n-1; i+=2) {
			
			if(i==n-1){
				a[0]=a[n-1];
				a[n-1]=1;
			}else{
				int t=a[i];
				a[i]=a[i+1];
				a[i+1]=t;
			}
		}
		
		for (int i = 0; i < a.length; i++) {
			out.print(a[i]+" ");
		}
		pr(ok(a,k));
//		pr(a);
	}
	boolean ok(int a[], int g){
		int x=0;
		for (int i = 0; i < a.length; i++) {
			if(gcd(a[i],i+1)>1)
				x++;
		}
		return x==g;
	}
	int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
	
	// IO methods
		
	void run() throws Exception {
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		pr(System.currentTimeMillis() - s + "ms");
	}
	public static void main(String[] args) throws Exception {new _361_B_Levko_and_Permutation().run();}

	InputStream in=System.in;
	PrintWriter out=new PrintWriter(System.out);
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = in.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {return !(c >= 33 && c <= 126);}
	private int skip() {int b;while ((b = readByte()) != -1 && isSpaceChar(b));return b;}

	public String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != // ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	public char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	public char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	public int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	
	
	public int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	public long nl(){
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	void pr(Object... ob) {if (!oj)System.out.println(Arrays.deepToString(ob).replace("],", "],\n"));}
}