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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String[] info = sc.nextLine().split(" ");
        int n = Integer.parseInt(info[0]);
        int k = Integer.parseInt(info[1]);
        sc.close();

        if (n >= k) {
            if (k == 1) {
                if (n == 1)
                    System.out.println("a");
                else
                    System.out.println(-1);
            }
            else {
                for (int i = 0; i < n - (k - 2); i++)
                    System.out.print(i % 2 == 0 ? 'a' : 'b');
                for (char c = 'c'; c < ('a' + k); c++)
                    System.out.print(c);
            }
        }
        else
            System.out.println(-1);
    }
}
