Little penguin Polo adores strings. But most of all he adores strings of length n.

One day he wanted to find a string that meets the following conditions:

  1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct. 
  2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n). 
  3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest. 



Help him find such string or state that such string doesn't exist.

String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.

Input

A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.

Output

In a single line print the required string. If there isn't such string, print "-1" (without the quotes).

Examples

Input

7 4


Output

ababacd


Input

4 7


Output

-1

import java.io.InputStreamReader ;
import java.io.BufferedReader ;
import java.io.IOException;



public class prob10 {

    public static void main(String[]args)throws IOException{
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            String[] in=br.readLine().split(" ");
            int n=Integer.parseInt(in[0]);
            int k=Integer.parseInt(in[1]);
            StringBuilder sb= new StringBuilder("");
            if(k==1&&n==1)
                System.out.println("a");
            else{
                if(k>n||(k==1 &&n!=1))
                System.out.println(-1);
            else{
                for(int i=0;i<n-k+2;i++)
                    sb=(i%2==0)?sb.append("a"):sb.append("b");
                int j=1;
                for(int i=n-k+2;i<n;i++){
                    sb.append((char)('b'+j));
                    j++;
                }
            }


            }
            System.out.println(sb);


        }


}
