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 PolothePenguin {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		if(k>n){
			System.out.println(-1);
			return;
		}
			
		if(k==1){
			if(n==1)System.out.println('a');
			else
				System.out.println(-1);
			return;
		}
		
		if(n>=2){
			StringBuilder sb = new StringBuilder("ab");
			int z = n-k;
			for(int i=0; i<z/2;i++)sb.append("ab");
			if(z%2==1)sb.append('a');
			int i=2;
			for(int j=0; j<k-2;j++){
				sb.append((char)('a'+i));
				i++;
			}
			System.out.println(String.valueOf(sb));
		}
		else{
			System.out.print('a');
		}
		
	}

}
