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 MaxN = 200005;
const double eps = 1e-7;
const double DINF = 1e100;
const int INF = 1000000006;
const long long LINF = 1000000000000000005ll;
string S[1000001];
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; i += 2) {
    S[i] = "a";
    S[i + 1] = "b";
  }
  if ((n > 1 && k < 2) || k > n) {
    cout << -1 << endl;
    return 0;
  }
  if (n % 2 != 0) S[n] = "a";
  string ks = "abcdefghijklmnopqrstuvwxyz";
  int i = n - 1;
  string res = "";
  for (i = n - 1; i > 0 && k > 2; i--) {
    S[i] = ks[--k];
    res = S[i] + res;
  }
  string r2 = "";
  if (k > 2) {
    cout << -1 << endl;
    return 0;
  }
  for (int j = 0; j <= i; j++) {
    r2 += S[j];
  }
  cout << r2 + res << endl;
  return 0;
}
