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;
char str[1000020];
int main() {
  int n, k, i, j, m;
  scanf("%d%d", &n, &k);
  if (n == 1 && k == 1)
    printf("a\n");
  else if (k > n || k == 1)
    printf("-1\n");
  else if (k == n) {
    for (i = 0; i < n; i++) {
      if (i == 0)
        str[0] = 'a';
      else
        str[i] = (char)((int)str[i - 1] + 1);
    }
    str[n] = '\0';
    printf("%s\n", str);
  } else if (k < n) {
    m = n - (k - 2);
    if (m % 2 == 0) {
      for (i = 0; i < m - 1; i = i + 2) {
        str[i] = 'a';
        str[i + 1] = 'b';
      }
      for (; i < n; i++) str[i] = (char)((int)str[i - 1] + 1);
    } else {
      for (i = 0; i < m - 1; i = i + 2) {
        str[i] = 'a';
        str[i + 1] = 'b';
      }
      str[i++] = 'a';
      str[i++] = 'c';
      for (; i < n; i++) str[i] = (char)((int)str[i - 1] + 1);
    }
    str[n] = '\0';
    printf("%s\n", str);
  }
  return 0;
}
