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;
long long left[3009], right[3009];
int main() {
  std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  string s, str = "";
  long long n, i, k;
  cin >> n >> k;
  if (k > n || (n > 1 && k == 1))
    cout << -1;
  else if (n == 1)
    cout << "a";
  else {
    long long t = k, tt = n;
    t -= 2;
    for (i = 0; i < n; i++) str += ('a' + (i & 1));
    n--, k--;
    while (t--) str[n--] = ('a' + k), k--;
    for (i = 0; i < tt; i++) cout << str[i];
  }
  cout << "\n";
  return 0;
}
