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.*;
import java.util.*;

public class Solution {
    
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();int k=sc.nextInt();
        if(k>n)
        {
            System.out.print(-1);
            return;
        }
      if(k==1)
      {
          if(n==1)
              System.out.print('a');
          else
              System.out.print(-1);
          return;
      }
        k-=2;
        int c=0;
        StringBuilder ans=new StringBuilder();
       while(n>0)
       {
           if(c%2==0)ans.append('a');
           else ans.append('b');
           c++;
           n--;
           if(n==k)
           {
               break;
           }
       }
        for(char cu='c';cu<='z';cu++)
        {
            if(k>0)
                ans.append(cu);
            else
                break;
            k--;
        }
        System.out.print(ans);
    }
}