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.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Optional;
import java.util.Scanner;
import java.util.stream.IntStream;

public class C361B {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        sc.close();

        PrintWriter out = new PrintWriter(System.out);
        Optional<int[]> result = solve(n, k);
        if (result.isPresent()) {
            out.println(Arrays.toString(result.get()).replaceAll("\\[|\\]|,", ""));
        } else {
            out.println(-1);
        }
        out.close();
    }

    private static Optional<int[]> solve(int n, int k) {
        if (k == n) return Optional.empty();
        int[] permutation = new int[n];
        permutation[0] = n - k;
        IntStream.rangeClosed(1, n - k - 1).forEach(i -> permutation[i] = i);
        IntStream.range(n - k, n).forEach(i -> permutation[i] = i + 1);
        return Optional.of(permutation);
    }
}