#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, a, b;
  cin >> n >> a >> b;
  string s;
  cin >> s;
  int f = -1;
  for (int i = 0; i * a <= n; i++) {
    int c = n - i * a;
    if (c % b == 0) {
      f = i;
      break;
    }
  }
  if (f == -1)
    cout << -1;
  else {
    int c = n - f * a;
    int d = c / b;
    int i, j = 0;
    cout << f + d << endl;
    for (i = 0; i < f; i++) {
      int k = j + a;
      for (j; j < k; j++) cout << s[j];
      cout << endl;
    }
    for (i = 0; i < d; i++) {
      int k = j + b;
      for (j; j < k; j++) cout << s[j];
      cout << endl;
    }
  }
}

### Prompt

Develop a solution in cpp 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;
int main() {
  int n, a, b;
  cin >> n >> a >> b;
  string s;
  cin >> s;
  int f = -1;
  for (int i = 0; i * a <= n; i++) {
    int c = n - i * a;
    if (c % b == 0) {
      f = i;
      break;
    }
  }
  if (f == -1)
    cout << -1;
  else {
    int c = n - f * a;
    int d = c / b;
    int i, j = 0;
    cout << f + d << endl;
    for (i = 0; i < f; i++) {
      int k = j + a;
      for (j; j < k; j++) cout << s[j];
      cout << endl;
    }
    for (i = 0; i < d; i++) {
      int k = j + b;
      for (j; j < k; j++) cout << s[j];
      cout << endl;
    }
  }
}

```