#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, p, q;
  string ch;
  cin >> n >> p >> q;
  cin >> ch;
  int x = 0, y = 0;
  bool r = false;
  if (n % p == 0) {
    x = n / p;
    r = true;
  } else if (n % q == 0) {
    y = n / q;
    r = true;
  } else {
    x = 1;
    while ((n - x * p) % q != 0 && x * p <= n) {
      x++;
    }
    if ((n - x * p) % q == 0) {
      y = (n - x * p) / q;
      r = true;
    } else {
      x = 0;
      while ((n - y * q) % p != 0 && y * q <= n) {
        y++;
      }
      if ((n - y * q) % p == 0) {
        x = (n - y * q) / p;
        r = true;
      }
    }
  }
  if (n - x * p - y * q != 0 || x < 0 || y < 0) {
    cout << -1 << endl;
  } else {
    cout << x + y << endl;
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < p; j++) {
        cout << ch[j + i * p];
      }
      cout << endl;
    }
    for (int i = 0; i < y; i++) {
      for (int j = 0; j < q; j++) {
        cout << ch[j + x * p + i * q];
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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, p, q;
  string ch;
  cin >> n >> p >> q;
  cin >> ch;
  int x = 0, y = 0;
  bool r = false;
  if (n % p == 0) {
    x = n / p;
    r = true;
  } else if (n % q == 0) {
    y = n / q;
    r = true;
  } else {
    x = 1;
    while ((n - x * p) % q != 0 && x * p <= n) {
      x++;
    }
    if ((n - x * p) % q == 0) {
      y = (n - x * p) / q;
      r = true;
    } else {
      x = 0;
      while ((n - y * q) % p != 0 && y * q <= n) {
        y++;
      }
      if ((n - y * q) % p == 0) {
        x = (n - y * q) / p;
        r = true;
      }
    }
  }
  if (n - x * p - y * q != 0 || x < 0 || y < 0) {
    cout << -1 << endl;
  } else {
    cout << x + y << endl;
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < p; j++) {
        cout << ch[j + i * p];
      }
      cout << endl;
    }
    for (int i = 0; i < y; i++) {
      for (int j = 0; j < q; j++) {
        cout << ch[j + x * p + i * q];
      }
      cout << endl;
    }
  }
  return 0;
}

```