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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class ZadB {
    public static void main(String... args) throws IOException {
        MyScanner sc = new MyScanner();
        int n = sc.nextInt();
        int k = sc.nextInt();
        int w = n-k;
        
        if (w==0){
            System.out.println("-1");
        } else {
            for (int i = 0; i < w-1; i++) {
                System.out.print((i+2) +" ");
            }
            System.out.print("1 ");
            for (int i = w; i < n; i++) {
                System.out.print((i+1) +" " );
            }
        }
    }

    static class MyScanner {

        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() throws IOException {
            if (st == null || !st.hasMoreTokens()) {
                String line = br.readLine();
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            String next = next();
            return Integer.parseInt(next);
        }
    }
}
