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;
long long gcd(long long a, long long b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
long long bigmod(long long n, long long power, long long mod) {
  if (!power) return 1;
  if (power % 2 == 0) {
    int ret = bigmod(n, power / 2, mod);
    return ((ret % mod) * (ret % mod)) % mod;
  } else
    return ((n % mod) * (bigmod(n, power - 1, mod) % mod)) % mod;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res *= x;
    x *= x;
    y >>= 1;
  }
  return res;
}
int n, k, fr2;
string s;
bool ok;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  if (k > n) return cout << "-1", 0;
  if (k == 1) {
    if (n == 1)
      cout << "a";
    else
      cout << "-1";
    return 0;
  }
  fr2 = n - k;
  s = "ab";
  ok = false;
  for (int i = 0; i < fr2; i++, ok = !ok) {
    s.push_back(char('a' + ok));
  }
  for (int i = 2; i < k; i++) s.push_back(char(i + 'a'));
  for (int i = 0; i < s.size() - 1; i++)
    if (s[i] == s[i + 1]) return cout << "-1", 0;
  cout << s << "\n";
  cin >> n;
  return 0;
}
