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.io.*;
import java.util.*;
public class LevkoandPermutation {
    public static void main(String[] args) throws Exception {
        BufferedReader k = new BufferedReader(new InputStreamReader(System.in));
        // Scanner k = new Scanner(System.in);

        //PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer s= new StringTokenizer(k.readLine());
        int n=Integer.parseInt(s.nextToken());
        int kk=Integer.parseInt(s.nextToken());
        if(n==kk){
            System.out.println(-1);
        }else{
            int l=n-kk-1;
            int a=kk+1;
            if(l%2==0){
                
                System.out.print(1);
                for (int i = 2; i <= a; i++) {
                    System.out.print(" "+i);
                }
                
                for (int i = a+1; i < n; i+=2) {
                    System.out.print(" "+(i+1)+" "+i);
                }
                System.out.println();
            }else{
                System.out.print(n);
                for (int i = 2; i <= a; i++) {
                    System.out.print(" "+i);
                }
                
                for (int i = a+1; i < n-1; i+=2) {
                    System.out.print(" "+(i+1)+" "+i);
                }
                System.out.println(" 1");
            }
        }
        
    }
}