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.*;
import java.io.*;
import java.math.*;

public class B {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int k = input.nextInt();

        if(n == k)  System.out.println(-1);
        else {
            if(k == 0) {
                for(int i = 1; i <= n - 1; i++) {
                    System.out.print(i + 1 + " ");
                }
                System.out.println(1);
                System.exit(0);
            }

            for(int i = 1; i <= n - k - 1; i++) {
                System.out.print(i + 1 + " ");
            }

            System.out.print(1 + " ");
            
            for(int i = n - k + 1; i < n; i++) {
                System.out.print(i + " ");
            }

            System.out.println(n);
        }   
    }
}
