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.*;
public class A288 {
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		int n = br.nextInt();
		int k = br.nextInt();
		if(k > n || (k == 1 && n > 1)){
			System.out.println(-1);
			return;
		}
		if(n == 1 && k == 1){
			System.out.println("a");
			return;
		}
		StringBuilder ans = new StringBuilder("");
		while(ans.length() < n-k+2){
			if(ans.length()%2 == 0){
				ans.append("a");
			}
			else{
				ans.append("b");				
			}
		}
		for(int i = 2;i<k;i++){
			ans.append((char)('a'+i));
		}
		System.out.println(ans);
	}
}
