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
n, k = map(int, raw_input().split())

letters = ['-', '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']

if n == 1 and k == 1:
    print 'a'
elif n < k or k <= 1:
    print -1
elif n == k:
    result = ''
    for i in range(n):
        result += letters[i + 1]
    print result
else:
    result = ''
    if n - k == 1:
        result = 'aba'
        left = n - 3
    elif k == 2:
        result = 'ab'*(n//2)
        if n%2 == 1:
            result += 'a'
        left = 0
    elif (n - k)%2 == 0:
      result = 'ab'*((n - k + 2)//2)
      left = k - 2
    else:
      result = 'ab'*((n - k + 1)//2) + 'a'
      left = k - 2


    for i in range(left):
        result += letters[i + 3]

    print result
        
