#include <bits/stdc++.h>
using namespace std;
template <int POS, class TUPLE>
void deploy(std::ostream &os, const TUPLE &tuple) {}
template <int POS, class TUPLE, class H, class... Ts>
void deploy(std::ostream &os, const TUPLE &t) {
  os << (POS == 0 ? "" : ", ") << get<POS>(t);
  deploy<POS + 1, TUPLE, Ts...>(os, t);
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::set<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T, class K>
std::ostream &operator<<(std::ostream &os, std::map<T, K> &make_pair) {
  int remain = make_pair.size();
  os << "{";
  for (auto e : make_pair)
    os << "(" << e.first << " -> " << e.second << ")"
       << (--remain == 0 ? "}" : ", ");
  return os;
}
using ll = long long;
ll n;
int k;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  if (n >= k) {
    ;
  } else {
    cout << -1 << "\n";
    return 0;
  }
  if (n == 1 && k == 1) {
    cout << 'a' << "\n";
    return 0;
  } else if (k == 1) {
    cout << -1 << "\n";
    return 0;
  }
  char now = 'a';
  string ans = "";
  for (int i = 0; i < n; i++) {
    ans += now;
    if (now == 'a') {
      now = 'b';
    } else {
      now = 'a';
    }
    if ((n - i - 1) == (k - 2)) {
      for (int j = 0; j < k - 2; j++) {
        ans += ('c' + j);
      }
      break;
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <int POS, class TUPLE>
void deploy(std::ostream &os, const TUPLE &tuple) {}
template <int POS, class TUPLE, class H, class... Ts>
void deploy(std::ostream &os, const TUPLE &t) {
  os << (POS == 0 ? "" : ", ") << get<POS>(t);
  deploy<POS + 1, TUPLE, Ts...>(os, t);
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::set<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T, class K>
std::ostream &operator<<(std::ostream &os, std::map<T, K> &make_pair) {
  int remain = make_pair.size();
  os << "{";
  for (auto e : make_pair)
    os << "(" << e.first << " -> " << e.second << ")"
       << (--remain == 0 ? "}" : ", ");
  return os;
}
using ll = long long;
ll n;
int k;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  if (n >= k) {
    ;
  } else {
    cout << -1 << "\n";
    return 0;
  }
  if (n == 1 && k == 1) {
    cout << 'a' << "\n";
    return 0;
  } else if (k == 1) {
    cout << -1 << "\n";
    return 0;
  }
  char now = 'a';
  string ans = "";
  for (int i = 0; i < n; i++) {
    ans += now;
    if (now == 'a') {
      now = 'b';
    } else {
      now = 'a';
    }
    if ((n - i - 1) == (k - 2)) {
      for (int j = 0; j < k - 2; j++) {
        ans += ('c' + j);
      }
      break;
    }
  }
  cout << ans << "\n";
  return 0;
}

```