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.Scanner;


public class PoloString { 
	
	
	boolean[] charsUsed;
	
	public static void main(String[] args) {
		new PoloString().solve();
	}
	
	
	private void solve() {
		charsUsed = new boolean[256];
		
		
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int k = s.nextInt();
		System.out.print(findString(n, k));
	}
	
	/**
	 * 
	 * @param n The length of the string
	 * @param k The number of unique characters
	 */
	private String findString(int n, int k) {
		if (n < k) {
			return "-1";
		}
		if (k == 1 && n > 1) {
			return "-1";
		}
		
		
		int i = 0;
		int length = 0;
		
		StringBuilder s = new StringBuilder();

		while (length < n) {
			
			s.append((char)(i + 'a'));
			length++;
			
			// if cannot repeat characters anymore
			if (n - length <= k - 1 ) {
				i++;
			}
			else {
				// swap between 0 and one
				i ^= 1;
			}
			// special case for when switching between states
			if (n - length == k && i == 1) {
				s.append((char)(1 + 'a'));
				s.append((char)(0 + 'a'));
				length += 2;
				i = 2;
			}
		}
		return s.toString();
	}
}
