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.PrintWriter;
import java.util.StringTokenizer;


public class Main {

    /**
     * @param args
     * @throws IOException 
     * @throws NumberFormatException 
     */
    public static void main(String[] args) throws NumberFormatException, IOException {
        
        
                
        BufferedReader r=new BufferedReader(new InputStreamReader(System.in));
        
    






StringTokenizer st=new StringTokenizer(r.readLine());
int n=Integer.parseInt(st.nextToken());
int k=Integer.parseInt(st.nextToken());
PrintWriter pr=new PrintWriter(System.out);
if(k>n||(n>k&&k==1))
{
    
    pr.println(-1);
}
    
else
    if(k==1&&n==1)
    {
        pr.println("a");
    }

else
{
int lim=n-(k-2);    
int x=lim/2;

for(int j=0;j<x;j++)
    {
    pr.print("ab");
}
if(lim%2!=0)
{

    pr.print("a");
}

for(int j=2;j<n-lim+2;j++)
{
    
    pr.print((char)('a'+j));
    
}

}
    pr.flush();
    pr.close();
    
    }
    

}
