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

public class LevkoAndPermutation {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        
        int N = sc.nextInt();
        int K = sc.nextInt();
        
        if (K == N) {
        	System.out.println(-1);
        	return;
        }
        
        int d = N - K;
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < d; i++) {
        	sb.append(((i + 1) % d) + 1);
        	sb.append(" ");
        }
        for (int i = d + 1; i <= N; i++) {
        	sb.append(i + " ");
        }
        System.out.println(sb.toString());
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

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

        int nextInt() {
            return Integer.parseInt(next());
        }
        
        int[] nextIntArray(int n) {
        	int[] arr = new int[n];
        	for (int i = 0; i < n; i++) {
        		arr[i] = nextInt();
        	}
        	return arr;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            String str = "";
            try { str = br.readLine(); }
            catch (IOException e) { e.printStackTrace(); }
            return str;
        }
    }
}