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

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author Igors
 */
public class CF177C {
    public static void main(String[] args) throws IOException{
        StreamTokenizer s= new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        s.nextToken();
        int n=(int)s.nval;
        s.nextToken();
        int k=(int)s.nval;
        long cur=1,last=-1;
        if (k>n){
            System.out.println(-1);
        }else if ((k==1) && (n>1)){
            System.out.println(-1);
        }else if ((k==1) && (n==1)){
            System.out.println('a');
        }        
        else {
            for(int i=0;i<n-k+2;i++){
                if (cur==1){
                    System.out.print('a');
                    cur=0;last=0;
                } else if (cur==0){
                    System.out.print('b');
                    cur=1;last=1;
                }
            }
            for(int i=1;i<=k-2;i++){
                System.out.print((char)('b'+i));
            }
            
        }
        
    }
}
