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;
const int inv = numeric_limits<int>::max();
const int minv = numeric_limits<int>::min();
const int max_n = 1000010;
int n, k;
char s[max_n + 10];
int main() {
  scanf("%d", &n);
  scanf("%d", &k);
  if (k == 1) {
    if (n == 1)
      printf("a\n");
    else
      printf("-1\n");
  } else {
    if (n < k)
      printf("-1\n");
    else {
      for (int i = n - k + 2; i <= n - 1; ++i) s[i] = (char('a' + i - n + k));
      s[0] = 'a';
      for (int i = 1; i <= n - k + 1; ++i) s[i] = (s[i - 1] == 'a' ? 'b' : 'a');
      s[n] = '\0';
      printf("%s\n", s);
    }
  }
  return 0;
}
