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.PrintWriter;
import java.util.Scanner;

public class Test {
	static int n, m, d;

	public static void main(String[] argv) {
		Scanner scan = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		String al = "abcdefghijklmnopqrstuvwxyz";
		int len = al.length();

		n = scan.nextInt();
		m = scan.nextInt();
		if (n < m || (n > 1 && m == 1)) {
			out.println("-1");
			out.close();
			return;
		}
		d = m;
		StringBuffer ret = new StringBuffer("");
		boolean usea = false, useb = false;
		for (int i = 0; i < n; ++i) {
			if (n - i == d) {
				int start = 0;
				if (!usea) {

				} else if (!useb) {
					start = 1;
				} else {
					start = 2;
				}
				for (int j = start; j < start + d; ++j)
					ret.append(al.charAt(j));
				break;
			} else {
				if (i % 2 == 0) {
					if (!usea)
						d--;
					usea = true;
					ret.append("a");
				} else {
					if (!useb)
						d--;
					useb = true;
					ret.append("b");
				}
			}
		}
		out.println(ret);
		out.close();
	}
}