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 ganaremos {

    public static void main(String[] args) {
        Scanner leer = new Scanner(System.in);
        int x = leer.nextInt(),j = leer.nextInt(),i;
        if(x < j || (x > 1 && j == 1)) {
            System.out.println(-1);
            return;
        }
        int[] s = new int[x];
        for (i = 0; i < x; i++){
            s[i] = i & 1;
        }
        int extra = j - 2;
        
        for (i = 0; i < extra; i++){
            s[x - extra + i] = i + 2;}
        
        char[] ans = new char[x];
        
        for (i = 0; i < x; i++){
            ans[i] = (char) (s[i] + 'a');
        }
        System.out.println(new String(ans));
    }

}