#include <bits/stdc++.h>
using namespace std;
long long n, m, i, j, t, ans, p, k, q;
int main() {
  ios_base::sync_with_stdio(0);
  static const size_t npos = -1;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  for (i = 0; i <= n / p + 1; i++) {
    for (j = 0; i * p + j * q <= n; j++) {
      if (i * p + j * q == n) {
        t = 1;
        break;
      }
    }
    if (t) {
      break;
    }
  }
  if (!t) {
    cout << -1 << endl;
    return 0;
  }
  cout << (i + j) << endl;
  for (k = 0, m = 0; m < i; k += p, m++) {
    cout << (s.substr(k, p)) << endl;
  }
  for (k = p * i, m = 0; m < j; k += q, m++) {
    cout << (s.substr(k, q)) << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, i, j, t, ans, p, k, q;
int main() {
  ios_base::sync_with_stdio(0);
  static const size_t npos = -1;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  for (i = 0; i <= n / p + 1; i++) {
    for (j = 0; i * p + j * q <= n; j++) {
      if (i * p + j * q == n) {
        t = 1;
        break;
      }
    }
    if (t) {
      break;
    }
  }
  if (!t) {
    cout << -1 << endl;
    return 0;
  }
  cout << (i + j) << endl;
  for (k = 0, m = 0; m < i; k += p, m++) {
    cout << (s.substr(k, p)) << endl;
  }
  for (k = p * i, m = 0; m < j; k += q, m++) {
    cout << (s.substr(k, q)) << endl;
  }
  return 0;
}

```