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.*;
public class Main
{
	public static void main(String[] args)
	{
		// get input
		// n, k

		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();

		// case 1
		// k > n
		// print -1 and exit
		if (k > n)
		{
			System.out.println(-1);
			return;
		}

		// set sb 
		StringBuilder sb = new StringBuilder();
		// case 2
		// k == n
		if (k == n)
		{
		// 
		// add a, b, c, .. k times
		// print it and exit

			for (int i = 0; i < k; i++)
			{
				sb.append((char)('a' + i));
			}

			System.out.println(sb.toString());
			return;
		}
		// case 3
		if (k == 1)
		{
			System.out.println(-1);
			return;

		}
		// case 4
		// otherwise
		// add a, b, a, b, ... (n - k + 2) times
		// and cdef... k-2 times
		for (int i = 0; i < n - k + 2; i++)
		{
			sb.append((char) ('a' + (i % 2)));
		}
		for (int i = 0; i < k - 2; i++)
		{
			sb.append((char) ('c' + i));
		}
		System.out.println(sb);
	}
}

