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() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  size_t n, p, q;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  vector<int> dp(n + 1);
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    if (i >= p) {
      dp[i] |= dp[i - p];
    }
    if (i >= q) {
      dp[i] |= dp[i - q];
    }
  }
  if (dp[n] == 0) {
    cout << -1;
    return 0;
  }
  size_t i = n;
  vector<string> answ;
  while (i > 0) {
    if (i >= p && dp[i - p] != 0) {
      answ.push_back(s.substr(i - p, p));
      i -= p;
    } else {
      answ.push_back(s.substr(i - q, q));
      i -= q;
    }
  }
  reverse(answ.begin(), answ.end());
  std::cout << answ.size() << endl;
  for (string s : answ) {
    cout << s << endl;
  }
  std::cout.flush();
  return 0;
}
