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.*;
import java.io.*;

public class C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int k = sc.nextInt();

		if (k > n) {
			System.out.println(-1);
		} else if (k == 1 && n != 1) {
			/* 1 letter and more than 1 length? Impossible */
			System.out.println(-1);
		} else if (k == 1 && n == 1) {
			System.out.println("a");
		} else {
			/* The filler will take care of a and b. */
			int filler = n - (k - 2);
			char[] ret = new char[n];
			for (int i = 0; i < filler; i++) {
				if (i % 2 == 0) {
					ret[i] = 'a';
				} else {
					ret[i] = 'b';
				}
				//ret = ret + (((i % 2) == 0) ? "a" : "b");
			}

			/* Do the rest of the letters. */
			for (int i = 0; i < (k - 2); i++) {
				ret[filler + i] = (char)('c' + i);
				//ret = ret + c;
			}
			System.out.println(ret);
		}
	}
}
