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 optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
void MAIN(ll tc) {
  int n, k;
  cin >> n >> k;
  if (n < k) {
    cout << -1;
    return;
  }
  if (n > 1 && k == 1) {
    cout << -1;
    return;
  }
  if (n == 1 && k == 1) {
    cout << 'a';
    return;
  }
  for (int i = 0; i < (n - k) / 2; i++) cout << "ab";
  if ((n - k) % 2) cout << 'a';
  if (n - k >= 1 && (n - k) % 2)
    cout << 'b';
  else
    cout << 'a';
  if (n - k >= 2 && (n - k) % 2)
    cout << 'a';
  else
    cout << 'b';
  for (int i = 2; i < k; i++) cout << char('a' + i);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    MAIN(i);
  }
}
