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 divp(int n, int p, string ch) {
  int i = p;
  int j = 0;
  while (i <= n) {
    while (j < i) {
      cout << ch[j];
      j++;
    }
    cout << endl;
    j = i;
    i += p;
  }
  return 0;
}
int divq(int n, int q, string ch) {
  int i = q;
  int j = 0;
  while (i <= n) {
    while (j < i) {
      cout << ch[j];
      j++;
    }
    cout << endl;
    j = i;
    i += q;
  }
  return 0;
}
int main() {
  int n, p, q, i, j;
  string ch, ch1;
  cin >> n >> p >> q;
  cin >> ch;
  if (!(n % p)) {
    cout << (n / p) << endl;
    divp(n, p, ch);
  } else if (!(n % q)) {
    cout << (n / q) << endl;
    divq(n, q, ch);
  } else if (!((n % p) % q)) {
    cout << (n / p) + ((n % p) / q) << endl;
    divp(((n / p) * p), p, ch);
    for (int j = ((n / p) * p); j <= n; j++) ch1.push_back(ch[j]);
    divq(n % p, q, ch1);
  } else {
    for (i = 0; (i <= n / p); i++) {
      if (!((n - i * p) % q)) {
        j = (n - i * p) / q;
        break;
      }
    }
    if ((i * p + q * j) > n)
      cout << "-1" << endl;
    else {
      cout << i + j << endl;
      divp((i * p), p, ch);
      for (int c = (i * p); c <= n; c++) ch1.push_back(ch[c]);
      divq((j * q), q, ch1);
    }
  }
  return 0;
}
