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;


public class LevkoPermutation {
    static Scanner input = new Scanner(System.in);
    public static void main(String[] args) {
        int n = input.nextInt(), K = input.nextInt();
        int bad = n - K;
        
        if (bad == 0)
            System.out.println("-1");
        else  {
            int[] perm = new int[n];
            for(int i=0;i<n;i++)
                perm[i] = i+1;
            int index = 0;
            if (bad % 2 == 1) {
                bad--; index++;
            }
            for(; bad > 0; index += 2, bad -= 2) {
                int temp = perm[index];
                perm[index] = perm[index+1];
                perm[index+1] = temp;
            }
            
            for(int i=0;i<n;i++)
                System.out.print(perm[i] + (i+1==n?"\n" : " "));
        }
    }
}
