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.*;
import java.io.*;
 
public class File {
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
		
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		
		int rem = n - k;
		
		
		
		if (k > n) {
			out.println(-1);
		}
		else if (k == 1) {
			if (n > 1) {
				out.println(-1);
			}
			else {
				out.println("a");
			}
		}
		else {
			StringBuilder sb = new StringBuilder();
			sb.append("ab");
			for (int i = 0; i < rem / 2; i++) {
				sb.append("ab");
			}
			if (rem % 2 == 1) {
				sb.append("a");
			}
			
			char c = 'c';
			for (int i = 0; i < k-2; i++) {
				sb.append(c);
				c++;
			}
			
			out.println(sb.toString());
		}
		
		
		out.close();
	}
}
