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.util.Scanner;

public class A {
    public static void main(String args[])
    {
        Scanner s=new Scanner(System.in);
        int n=s.nextInt(),k=s.nextInt();
        String result=null;
        if(n<k){
            result="-1";
        }else{
            if(n==1)result="a";
            else{
                if(k==1)result="-1";
                else{
                    char a='a',b='b';
                    StringBuffer sb=new StringBuffer();
                    int r=k-2,i=0;
                    for(;i<n-r;i+=2)sb.append(a).append(b);
                    if(i!=n-r)sb.deleteCharAt(n-r);
                    i=0;
                    while(i++!=r)sb.append((char)(b+i));
                    result=sb.toString();
                }
            }
            
        }
        System.out.println(result.toString());
        s.close();

    }
}
