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;
class ProblemATheTextSplitting {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int n, p, q;
    in >> n >> p >> q;
    string s;
    in >> s;
    int a = 0, b = 0;
    for (; a <= n; ++a) {
      for (b = 0; b <= n; ++b) {
        if (a * p + b * q > n) break;
        if (a * p + b * q == n) {
          out << a + b << '\n';
          int pos = 0;
          while (a--) out << s.substr(pos, p) << '\n', pos += p;
          while (b--) out << s.substr(pos, q) << '\n', pos += q;
          return;
        }
      }
    }
    out << -1;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ProblemATheTextSplitting solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
