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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int main() {
  int n, k;
  cin >> n >> k;
  if ((n > 1 && k == 1) || k > n)
    cout << "-1\n";
  else {
    int x = 97;
    if (k == 1)
      cout << char(x) << endl;
    else if (k == 2) {
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << char(x);
        else
          cout << char(x + 1);
      }
      cout << endl;
    } else {
      int i;
      for (i = 0; i < n - (k - 2); i++) {
        if (i % 2 == 0)
          cout << char(x);
        else
          cout << char(x + 1);
      }
      x += 2;
      for (; i < n; i++) {
        cout << char(x);
        x++;
      }
      cout << endl;
    }
  }
  return 0;
}
