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;
int main() {
  int n, k;
  int us[40];
  while (scanf("%d%d", &n, &k) == 2) {
    int qus = 0;
    if (n == 1 && k == 1)
      printf("a\n");
    else if (n < k || k == 1)
      printf("-1\n");
    else {
      memset(us, 0, sizeof(us));
      string res;
      for (int i = 0; i < (n); i++) {
        char ch;
        if (n - i <= k - qus)
          ch = 'a' + qus;
        else if (i % 2 && k > 1)
          ch = 'b';
        else
          ch = 'a';
        if (!us[ch - 'a']) qus++;
        us[ch - 'a'] = 1;
        res += ch;
      }
      if (qus != k)
        printf("-1\n");
      else
        cout << res << endl;
    }
  }
  return 0;
}
