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.HashSet;
import java.util.*; 
import java.util.Scanner;
import java.util.Set;
import java.util.ArrayList;
import java.util.Collections;
import java.text.DecimalFormat;
import java.lang.Math; 
public class Lo {
    public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int m = sc.nextInt();
      if(n==m){
          System.out.println(-1);
      }
      else{
          for(int i = 1; i <= n-m; i++){
              if(i==1){
                  System.out.print((n-m)+" ");
              }
              else{
                  System.out.print((i-1)+" ");
              }
          }
          for(int i = n-m+1; i <= n; i++){
              System.out.print(i+" ");
          }
      }
    }
}