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;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  vector<string> result;
  int c_p = 0, c_q = 0, t = n;
  bool is_ok = false;
  while (!is_ok) {
    t = n;
    c_q = 0;
    t -= c_p * p;
    if (t < 0) break;
    while (t > 0) {
      t = n;
      c_q++;
      t -= c_p * p + c_q * q;
    }
    if (t == 0)
      is_ok = true;
    else
      c_p++;
  }
  if (is_ok) {
    while (c_p > 0) {
      result.push_back(s.substr(0, p));
      s.erase(0, p);
      c_p--;
    }
    while (c_q > 0) {
      result.push_back(s.substr(0, q));
      s.erase(0, q);
      c_q--;
    }
  }
  if (result.size()) {
    cout << result.size() << endl;
    for (int i = 0; i < result.size(); i++) cout << result[i] << endl;
  } else
    cout << -1 << endl;
}
