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 C {
	public static void main(String [] args){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		StringBuffer s = new StringBuffer("");
		if(k>n){
			System.out.println(-1);
			return ;
		}
		if(k == 1 && n > 1){
			System.out.println(-1);
			return;
		}
		if(k == 1 && n==1){
			System.out.println("a");
			return;
		}
		int l = n - k + 2;
		boolean t = true ;
		for(int i = 0 ; i < l ; i++){
			if(t)
			s.append('a');
			else
				s.append('b');
			t=!t;
		}
		char c= 'b';
		for(int i = 0 ; i < k - 2 ; i++){
			c++;
			s.append(c);
		}
		System.out.println(s);
	}
}
