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 math
import itertools
import collections

def getdict(n):
    d = {}
    if type(n) is list:
        for i in n:
            if i in d:
                d[i] += 1
            else:
                d[i] = 1
    else:
        for i in range(n):
            t = ii()
            if t in d:
                d[t] += 1
            else:
                d[t] = 1
    return d
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def prime(n):
    if n == 2:
        return True
    if n % 2 == 0 or n <= 1:
        return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0:
            return False
    return True
def revn(n):
    m = 0
    while n > 0:
        m = m * 10 + n % 10
        n = n // 10
    return m

n, k = mi()
sample = 'abcdefghijklmnopqrstuvwxyz'
if n == k:
    print(sample[:k])
elif n < k or n > 1 and k == 1:
    print(-1)
else:
    print('ab' * ((n - k + 2) // 2) + 'a' * ((n - k) % 2) + sample[2:k])
