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 buf[2000001];
int main() {
  int n, k;
  int diffcnt = 0;
  int used[26] = {};
  scanf("%d%d", &n, &k);
  if (n == 1 && k == 1) {
    puts("a");
  } else if (k == 1) {
    puts("-1");
  } else {
    for (int i = 1; i <= n; i++) {
      int j;
      for (j = 0; j < k; j++) {
        if (buf[i - 1] != j + 'a') {
          bool ch = false;
          if (!used[j]) {
            ch = true;
            used[j] = 1;
            diffcnt++;
          }
          if (n - i - k + diffcnt >= 0) {
            buf[i] = j + 'a';
            break;
          }
          if (ch) {
            used[j] = 0;
            diffcnt--;
          }
        }
      }
      if (j == k) {
        printf("-1\n");
        return 0;
      }
    }
  }
  puts(&buf[1]);
}
