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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class A {

	public static void main(String[] args) throws IOException {
		PrintWriter out = new PrintWriter(System.out);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		StringTokenizer st = new StringTokenizer(br.readLine());

		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());

		StringBuilder sb = new StringBuilder();

		if (n == k) {
			for (int i = 0; i < n; i++) {
				sb.append((char) ('a' + i));
			}
		} else if (k > n || (k == 1 && n > 1)) {
			sb.append("-1");
		} else {
			boolean a = true;
			for (int i = 0; i < n - k + 2; i++) {
				if (a)
					sb.append("a");
				else
					sb.append("b");
				a = !a;
			}
			for (int i=2; i<k; i++)
				sb.append((char) ('a' + i));
		}
		
		out.println(sb.toString());
		
		br.close();
		out.close();
	}
}
