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.*;

public class PoloThePenguinAndStrings {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int k = cin.nextInt();
        cin.close();
        if (k > n || (n > 1 && k == 1)) {
            System.out.print("-1");
            System.exit(0);
        } else if (n == 1 && k == 1) {
            System.out.println("a");
        } else {
            char sol[] = new char[n];
            int i, j;
            for (i = n - 1; k > 2; i--, k--) {
                sol[i] = (char) ('a' + k - 1);
            }
            for (j = 0; j < i; j += 2) {
                sol[j] = 'a';
                sol[j + 1] = 'b';
            }
            if (j == i)
                sol[j] = 'a';
            System.out.println(sol);
        }
    }
}
