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
#include <bits/stdc++.h>
using namespace std;
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<T> &v) {
  for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
    out << v[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<pair<T, T> > &v) {
  for (size_t i = 0; i < v.size(); ++i)
    out << "(" << v[i].first << ", " << v[i].second << ") ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<vector<T> > &v) {
  for (size_t i = 0; i < v.size(); ++i) {
    for (size_t j = 0; j < v[i].size(); ++j) {
      out << v[i][j] << " ";
    }
    out << "\n";
  }
  return out;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  if (k > n || (n > 1 && k == 1)) {
    cout << "-1\n";
    return 0;
  }
  char x = 'a', y = 'b';
  string ans = "a";
  --n;
  while (n) {
    if (ans.back() == x) {
      ans.push_back(y);
    } else {
      ans.push_back(x);
    }
    --n;
  }
  k -= 2;
  int i = 0;
  while (i < k) {
    ans.pop_back();
    ++i;
  }
  x = 'c';
  for (int i = n - k; i < n; ++i) {
    ans.push_back(x);
    ++x;
  }
  cout << ans << "\n";
  return 0;
}
