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 PoloPenguinAndString {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		while (s.hasNext())
		{
			int n = s.nextInt(), k = s.nextInt();
			if (n==1 && k==1)
			{
				System.out.println("a");
				continue;
			}
			
			if (k > n || k ==1)
			{
				System.out.println(-1);
				continue;
			}
//			int remain = n, used = 0;
//			boolean ona = true;
//			for ( ;remain != k - used && remain > 0; remain--)
//			{
//				if (ona)
//				{
//					ans += "a";
//					used = Math.max(used, 1);
//					ona = false;
//				}
//				else
//				{
//					ans += "b";
//					used = Math.max(used, 2);
//					ona = true;
//				}
//			}
			int totAbs = (n - k + 2);
			int pairAbs = totAbs / 2;
			boolean extra = totAbs % 2 != 0;
			for (int i = 0;i<pairAbs;i++)
				System.out.print("ab");
			if (extra)
				System.out.print("a");
			
			k-=2;
			for (int i=0 ;i < k; i++)
			{
				System.out.print((char) ('a' + (i + 2)));
			}
			System.out.println();
		}
	}

}
