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;
void substr(int start, int end, char in[], char out[]) {
  for (int i = start; i < end; ++i) out[i - start] = in[i];
  out[end - start] = '\0';
}
int main() {
  int n, p, q;
  char s[100];
  cin >> n >> p >> q;
  cin >> s;
  int pdivmax = n / p + 1, qdivmax = n / q + 1;
  bool divided = false;
  if (p == q) {
    if (n % p == 0) {
      cout << n / p << endl;
      for (int i = 0; i < n; i += p) {
        char out[100];
        substr(i, i + p, s, out);
        cout << out << endl;
      }
      divided = true;
    }
  } else {
    for (int i = 0; i < pdivmax; ++i) {
      for (int j = 0; j < qdivmax; ++j) {
        if (p * i + q * j == n) {
          cout << i + j << endl;
          for (int k = 0; k < p * i; k += p) {
            char out[100];
            substr(k, k + p, s, out);
            cout << out << endl;
          }
          for (int k = p * i; k < n; k += q) {
            char out[100];
            substr(k, k + q, s, out);
            cout << out << endl;
          }
          divided = true;
          break;
        }
      }
      if (divided) break;
    }
  }
  if (!divided) cout << -1 << endl;
  return 0;
}
