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.*;
import java.util.*;

public class B {
	public static void main(String[] args) {
		FastReader in = new FastReader();
		
		int n = in.nextInt();
		int k = in.nextInt();

		if(k == n) System.out.print(-1);
		else {
			System.out.print(n-k+" ");
			for(int i = 1; i < n-k; i++)
				System.out.print( i + " ");
			for(int i = n-k+1; i<=n; i++)
				System.out.print( i + " ");
		}
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		public FastReader() { 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;
		}
		int[] readArray(int size) {
			int[] arr = new int[size];
			for(int i = 0; i < size; i++)
				arr[i] = Integer.parseInt(next());
			return arr;
		}
		int[][] read2dArray(int rows, int cols) {
			int[][] arr = new int[rows][cols];
			for(int i = 0; i < rows; i++) {
				for(int j = 0; j < cols; j++)
					arr[i][j] = Integer.parseInt(next());
			}
			return arr;
		}
	}
}