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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.PriorityQueue;


public class B {
	public static void main(String[] args) {
		MScanner sc = new MScanner();
		PrintWriter out = new PrintWriter(System.out);
		int N = sc.nextInt();
		int K = sc.nextInt();
		if(N==K)out.println("-1");
		else{
			out.print(N-K);
			for(int a=1;a<N-K;a++)out.print(" "+a);
			for(int a=N-K+1;a<=N;a++)out.print(" "+a);
		}
		out.close();
	}



	static class MScanner{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		
		public MScanner(){
			stream = System.in;
			//stream = new FileInputStream(new File("dec.in"));

		}
		int read(){
			if(numChars==-1)
				throw new InputMismatchException();
			if(curChar>=numChars){
				curChar = 0;
				try{
					numChars = stream.read(buf);
				} catch (IOException e){
					throw new InputMismatchException();
				}
				if(numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
		
		boolean isSpaceChar(int c){
			return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
		}
		boolean isEndline(int c){
			return c=='\n'||c=='\r'||c==-1;
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		int[] nextIntArray(int N){
			int[] ret = new int[N];
			for(int a=0;a<N;a++)
				ret[a] = nextInt();
			return ret;
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		long[] nextLongArray(int N){
			long[] ret = new long[N];
			for(int a=0;a<N;a++)
				ret[a] = nextLong();
			return ret;
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		double[] nextDoubleArray(int N){
			double[] ret = new double[N];
			for(int a=0;a<N;a++)
				ret[a] = nextDouble();
			return ret;
		}
		String next(){
			int c = read();
			while(isSpaceChar(c))
				c=read();
			StringBuilder res = new StringBuilder();
			do{
				res.appendCodePoint(c);
				c=read();
			} while(!isSpaceChar(c));
			return res.toString();
		}
		String[] nextArray(int N){
			String[] ret = new String[N];
			for(int a=0;a<N;a++)
				ret[a] = next();
			return ret;
		}
		String nextLine(){
			int c = read();
			while(isEndline(c))
				c=read();
			StringBuilder res = new StringBuilder();
			do{
				res.appendCodePoint(c);
				c = read();
			}while(!isEndline(c));
			return res.toString();
		}
		
	}
}
