You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input

The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output

If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Examples

Input

5 2 3
Hello


Output

2
He
llo


Input

10 9 5
Codeforces


Output

2
Codef
orces


Input

6 4 5
Privet


Output

-1


Input

8 1 1
abacabac


Output

8
a
b
a
c
a
b
a
c
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, p, q;
  string s;
  bool can = true;
  vector<string> vec;
  int aux = 0;
  int pos = 0;
  cin >> n >> p >> q;
  cin >> s;
  aux = n;
  if (n % p == 0) {
    cout << n / p << endl;
    for (int i = 0; i < n; i += p) {
      cout << s.substr(i, p) << endl;
    }
  } else if (n % q == 0) {
    cout << n / q << endl;
    for (int i = 0; i < n; i += q) {
      cout << s.substr(i, q) << endl;
    }
  } else {
    while (((aux % q != 0)) && pos < n) {
      vec.push_back(s.substr(pos, p));
      pos += p;
      aux -= p;
      if (aux < 0) {
        can = false;
      }
    }
    if (can) {
      while (pos < n) {
        vec.push_back(s.substr(pos, q));
        pos += q;
      }
      cout << vec.size() << endl;
      for (int i = 0; i < vec.size(); i++) {
        cout << vec[i] << endl;
      }
    } else {
      cout << -1;
    }
  }
  return 0;
}
