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.io.PrintWriter;
import java.util.Scanner;



public class C {
	public  static  Scanner scan = new Scanner(System.in);
	
	
	public static void solve () {
		int n=scan.nextInt();
		int k=scan.nextInt();
		 PrintWriter out = new PrintWriter(System.out);
		if(k>n) out.println("-1");
		else if(k==1){
			if(n==1) out.print	("a");
			else out.println("-1");
		}
		else {
			int x= n-k+2;
			//System.out.println(x);
			for(int i=0;i<x/2;i++) out.print("ab");
			if(x%2==1) out.print("a");
			for(int i=0;i<k-2;i++) out.print((char)('c'+i));
		}
		out.close();
	}
	public static void main(String[] args) {
	
		solve();
		scan.close();
	}

}
