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 A288 {

	static Scanner in = new Scanner(System.in);
	
	public static void main(String[] args) {
		char[] wrd = {'a',
				'b',
				'c',
				'd',
				'e',
				'f',
				'g',
				'h',
				'i',
				'j',
				'k',
				'l',
				'm',
				'n',
				'o',
				'p',
				'q',
				'r',
				's',
				't',
				'u',
				'v',
				'w',
				'x',
				'y',
				'z'
		};
		int n = in.nextInt();
		int k = in.nextInt();
		int a = k;
		if ((n >= k && k <= 26) && ((k == 1 && n == 1) || (k != 1 && n > 1)) ){
		
				while(n-a > 0){
					System.out.print("a"); 
					a+=2;
					if (n - a >= 0) System.out.print("b");
				}

				
				if ((n-k) % 2 == 0){
					for (int j = 0; j < k; j++) {
						System.out.print(wrd[j]);
					}
				} else {
					System.out.print("b");
					System.out.print("a");
					for (int j = 0; j < k-2; j++) {
						System.out.print(wrd[j+2]);
					}
				}
						
		} else System.out.println(-1);
		
	}

}
