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.Arrays;
import java.util.Scanner;

public class ProblemA {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = in.nextInt();
		int k = in.nextInt();
		
		if (n < k || (k == 1 && n >= 2)) {
			out.println(-1);
			out.flush();
		} else {
			StringBuffer b = new StringBuffer();
			for (int i = 0 ; i < n ; i++) {
				if (i % 2 == 0) {
					b.append('a');
				} else {
					b.append('b');
				}
			}
			char[] cl = b.toString().toCharArray();
			int ci = n - 1;
			while (k >= 3) {
				cl[ci] = (char)('a' + (k - 1));
				ci--;
				k--;
			}
			out.println(cl);
		}
		out.flush();
	}
	
	public static void debug(Object... o) {
		System.err.println(Arrays.deepToString(o));
	}
}
