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 used[10007];
int pr[10007];
int main() {
  int n, p, q, x;
  std::cin >> n >> p >> q;
  std::string str, str2;
  std::cin >> str;
  std::vector<std::string> ans;
  pr[0] = -1;
  std::queue<int> queue;
  queue.push(0);
  while (!queue.empty()) {
    x = queue.front();
    queue.pop();
    if (x == n) {
      for (int i = x; i != 0;) {
        p = pr[i];
        q = i - p;
        str2 = "";
        for (int j = 0; j < q; j++) {
          str2.push_back(str[0]);
          str.erase(0, 1);
        }
        ans.push_back(str2);
        i = p;
      }
      std::cout << ans.size() << std::endl;
      for (auto &s : ans) std::cout << s << std::endl;
      return 0;
    }
    if (x + q <= n && !used[x + q]) {
      used[x + q] = 1;
      pr[x + q] = x;
      queue.push(x + q);
    }
    if (x + p <= n && !used[x + p]) {
      used[x + p] = 1;
      pr[x + p] = x;
      queue.push(x + p);
    }
  }
  std::cout << -1;
  return 0;
}
