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>
int main() {
  int n, p, q;
  std::cin >> n >> p >> q;
  std::string s;
  std::cin >> s;
  std::vector<int> decomp(n + 1, -1);
  decomp[0] = 0;
  for (int k = 0; k < n; k++) {
    if (decomp[k] < 0) {
      continue;
    }
    if (k + p <= n) {
      decomp[k + p] = p;
    }
    if (k + q <= n) {
      decomp[k + q] = q;
    }
  }
  if (decomp[n] < 0) {
    puts("-1");
    return 0;
  }
  std::vector<int> lengths;
  int index(n);
  while (index > 0) {
    lengths.push_back(decomp[index]);
    index -= decomp[index];
  }
  std::cout << lengths.size() << std::endl;
  index = 0;
  for (int p = 0; p < lengths.size(); p++) {
    std::cout << s.substr(index, lengths[p]) << std::endl;
    index += lengths[p];
  }
  return 0;
}
