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 Main2 {
    public static void main(String args[]){
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int k = input.nextInt();

        System.out.println(QR(n,k));
    }
    
    public static String QR(int n,int k){
    	if(k == 1){
    		if(n == 1){
    			return "a";
    		}else{
    			return "-1";
    		}
    	}
    	if(k > n)return "-1";
    	int stnum = 99;
    	StringBuilder sb = new StringBuilder();
    	for(int i = 0 ; i < n-(k-2) ; i++){
    		if(i % 2 == 0){
    			sb.append("a");
    		}else{
    			sb.append("b");
    		}
    	}
    	for(int i = n-(k-2) ; i < n ; i++){
    		sb.append((char)(stnum+i-(n-(k-2))));
    	}
    	return sb.toString();
    }
}