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
function trim(s) {
	return s.replace(/^\s+|\s+$/gm, '');
}

function tokenize(s) {
	return trim(s).split(/\s+/);
}

function tokenizeIntegers(s) {
	var tokens = tokenize(s);
	for (var i = 0; i < tokens.length; i += 1) {
		tokens[i] = parseInt(tokens[i]);
	};
	return tokens;
}

function main() {
	var integers = tokenizeIntegers(readline());
	var length = integers[0], variety = integers[1];

	if ((variety == 1 && length > 1) || (variety > length)) {
		print(-1);
		return;
	}

	var tailLength = Math.max(0, variety-2);
	var headLength = Math.max(length-tailLength);
	var tailStartCharCode = 'c'.charCodeAt(0);
	var characters = [];
	for (var i = 0; i < headLength; i += 1) {
		characters.push(i%2 == 0 ? 'a' : 'b');
	}
	for (var i = 0; i < tailLength; i += 1) {
		characters.push(String.fromCharCode(tailStartCharCode+i));
	}

	print(characters.join(''));
}

main();
