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 CF177C {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int k = scan.nextInt();
		if(k > n){
			System.out.println("-1");
			return;
		}
		if(k == 1 && n > 1){
			System.out.println("-1");
			return;
		}
		if(n == 1){
			System.out.println("a");
			return;
		}
		StringBuilder ans = new StringBuilder("");
		char start = (char) ('a'+k-1);
		for(int i = 0; i < k-2; i++){
			ans.append(start+"");
			start--;
		}
		int count = 0;
		for(int i = k-2; i < n; i++){
			if((n - k) % 2 == 1){
				if(count % 2 == 0){
					ans.append("a");
				}
				else{
					ans.append("b");
				}
			}
			else{
				if(count % 2 == 0){
					ans.append("b");
				}
				else{
					ans.append("a");
				}
			}
			count++;
		}
		System.out.println(ans.reverse());
	}
}