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:64000000")
using namespace std;
template <typename _T>
_T Max(_T a, _T b) {
  return a < b ? b : a;
}
template <typename _T>
_T Max(_T a, _T b, _T c) {
  return a < Max(b, c) ? Max(b, c) : a;
}
template <typename _T>
_T Min(_T a, _T b) {
  return a > b ? b : a;
}
template <typename _T>
_T Min(_T a, _T b, _T c) {
  return a > Min(b, c) ? Min(b, c) : a;
}
template <typename _T>
_T Abs(_T a) {
  return (a > 0) ? (a) : (-a);
}
const int inf = -(1 << 29);
const int INF = (1 << 29);
const int M = 105;
int mat[M * M];
int main() {
  int len, dis;
  while (cin >> len >> dis) {
    if (dis > len) {
      cout << -1 << endl;
    } else if (len == 1) {
      cout << "a" << endl;
    } else if (dis == 1) {
      cout << "-1" << endl;
    } else {
      int ab = len - dis + 2;
      for (int i = 0; i < ab / 2; i++) {
        cout << "ab";
      }
      (ab & 1) ? (cout << "a") : (cout << "");
      for (int i = 0; i < len - ab; i++) {
        printf("%c", 'c' + i);
      }
      cout << endl;
    }
  }
  return 0;
}
