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.util.*;
public class Main {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();
        if (k >= n){
            System.out.println(-1);
            return;
        }
        if ((n-k) % 2 == 0){
            StringBuilder fin = new StringBuilder();
            int last = 0;
            for (int i = 0; i < (n - k); i++) {
                if ((i&1)==0){
                    fin.append(i+2+" ");
                }
                else {
                    fin.append(i+" ");
                }
                last = i;
            }
            for (int i = last + 1; i< n; i++){
                fin.append(i+1+" ");
            }
            System.out.println(fin);
        }
        else {
            StringBuilder fin = new StringBuilder();
            fin.append(1+" ");
            int last = 0;
            for (int i = 1;i < (n - k);i++){
                if ((i&1)==1){
                    fin.append(i+2+" ");
                }
                else {
                    fin.append(i+" ");
                }
                last = i;
            }
            for (int i = last + 1; i< n; i++){
                fin.append(i+1+" ");
            }
            System.out.println(fin);
        }
        
    }
}