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.Scanner;

/**
 * Created with IntelliJ IDEA.
 * User: orhan
 */
public class B {
    public static void main(String [] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();

        int [] a = new int[n+1];
        for(int i = 1; i<=n; i++) a[i] = i;

        if(n == k) System.out.println(-1);
        else {
            int change = n-1-k;
            int count = (change)/2;

            for(int i = 1; i<=count; i++) {
                int ind = 2*i;
                a[ind] = a[ind] ^ a[ind+1];
                a[ind+1] = a[ind] ^ a[ind+1];
                a[ind] = a[ind] ^ a[ind+1];
            }

            if(change % 2 == 1) {
                a[1] = a[1] ^ a[n];
                a[n] = a[1] ^ a[n];
                a[1] = a[1] ^ a[n];
            }

            for(int i = 1; i<=n; i++) {
                System.out.print(a[i]);
                if(i != (n)) System.out.print(" ");
            }

            System.out.println();
        }

    }
}
