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() {
  int tam;
  int p;
  int q;
  int cont = 0;
  string frase;
  vector<string> respuesta;
  cin >> tam >> p >> q;
  cin >> frase;
  if (p + q == tam) {
    respuesta.push_back(frase.substr(0, p));
    respuesta.push_back(frase.substr(p, tam - 1));
    cont += 2;
  } else {
    if ((tam % q == 0 || tam % p == 0)) {
      if (tam % q == 0) {
        while (tam > 0) {
          respuesta.push_back(frase.substr(0, q));
          frase.erase(0, q);
          tam -= q;
          cont++;
        }
      } else if (tam % p == 0) {
        while (tam > 0) {
          respuesta.push_back(frase.substr(0, p));
          frase.erase(0, p);
          tam -= p;
          cont++;
        }
      }
    } else if (p + q > tam && (tam % q != 0 || tam % p != 0)) {
      respuesta.push_back("-1");
    } else if (p + q < tam) {
      int mayor = (p > q ? p : q);
      int menor = (p < q ? p : q);
      while (tam > 0 && tam >= mayor && tam % menor != 0) {
        respuesta.push_back(frase.substr(0, mayor));
        frase.erase(0, mayor);
        tam -= mayor;
        cont++;
      }
      while (tam > 0 && menor <= tam) {
        respuesta.push_back(frase.substr(0, menor));
        frase.erase(0, menor);
        tam -= menor;
        cont++;
      }
      if (tam > 0) {
        respuesta.erase(respuesta.begin(), respuesta.end());
        respuesta.push_back("-1");
        cont = 0;
      }
    }
  }
  if (respuesta[0] != "-1") {
    cout << cont << endl;
  }
  for (int i = 0; i < respuesta.size(); i++) {
    cout << respuesta[i] << endl;
  }
  return 0;
}
