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 static java.lang.Math.*; 

import java.io.*; 
import java.util.*; 
import java.math.*;

public class Template {
	BufferedReader in; 
	PrintWriter out;
	StringTokenizer st; 

	String next() {
		while (st==null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(in.readLine());
			} catch (Exception e) {}
		}
		return st.nextToken(); 
	}

	int nextInt() {
		return Integer.parseInt(next()); 
	}

	long nextLong() {
		return Long.parseLong(next());
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

	public void run() throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		Long start = System.currentTimeMillis();
		
		int n = nextInt(); 
		int k = nextInt(); 

		if (n < k || (n>1 && k==1)) {
			out.println(-1);
		} else {
			if (n==k) {
				for (int i=0; i<n; i++) out.print((char)('a' + i));
			} else {
				for (int i=0; i<n-(k-2); i++) {
					out.print((char)('a' + i % 2));
				}
				for (int i=2; i<k; i++) out.print((char)('a'+i));
			}
		}

		Long end = System.currentTimeMillis();

		//System.out.println((end - start) / 1000.0);
		out.close(); 
	}
	

	public static void main(String[] args) throws Exception {
		new Template().run(); 
	}
}