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>
#pragma comment(linker, "/STACK:60000000")
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-12;
const int INF = (1 << 31) - 1;
const long long LLINF = ((long long)1 << 62) - 1;
set<char> s;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (k > n) {
    printf("-1");
    return 0;
  }
  if (k == 1 && n > 1) {
    printf("-1");
    return 0;
  }
  string res = "";
  int t = 0;
  for (char c = 'a'; c <= 'z'; c++) s.insert(c);
  for (int i = 0; i < n; i++) {
    if (k == n - i) {
      res += *s.begin();
      s.erase(s.begin());
      k--;
    } else {
      if (!t) {
        res += 'a';
        s.erase('a');
      } else {
        res += 'b';
        s.erase('b');
      }
      t = 1 - t;
      if ((int)res.size() == 1 || (int)res.size() == 2) k--;
    }
  }
  puts(res.c_str());
  return 0;
}
