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>
int main() {
  int i, d, n, k;
  char str[1000005], str2[1000005];
  scanf("%d%d", &n, &k);
  if ((k > n) || ((k == 1) && (n != 1))) {
    printf("-1");
  } else {
    if ((k == 1) && (n == 1)) {
      printf("a");
    } else {
      for (i = 1; i <= k; i++) {
        str[i] = 97 + (i - 1);
      }
      d = n - k;
      for (i = 1; i <= d; i++) str2[i] = 97 + ((i - 1) % 2);
      for (i = 1; i <= 2; i++) printf("%c", str[i]);
      for (i = 1; i <= d; i++) printf("%c", str2[i]);
      for (i = 3; i <= k; i++) printf("%c", str[i]);
    }
  }
  return 0;
}
