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.util.*;

public class B {
    static int GCD(int a, int b) {
        if (b == 0)
            return a;
        return GCD(b, a % b);
    }

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        int n = s.nextInt();
        int k = s.nextInt();
        int[] arr = new int[n + 1];

        if (k > n - 1) {
            System.out.println(-1);
        } else {
            for (int i = 0; i < k; i++) {
                arr[arr.length - i - 1] = arr.length - i - 1;
            }
            arr[arr.length - k - 1] = 1;
            int count = 2;
            for (int i = 1; i < arr.length - k - 1; i++)
                arr[i] = count++;

            for (int i = 1; i < arr.length; i++)
                System.out.print(arr[i] + " ");
        }
    }
}