#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const int N = 1e5 + 3;
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void set_bit(int& n, int b) { n |= (1ll << (b)); }
inline void unset_bit(int& n, int b) { n &= ~(1ll << (b)); }
int main() {
  int n, p, q;
  int x, y;
  string s;
  cin >> n >> p >> q;
  cin >> s;
  x = y = -1;
  for (int i = 0; i <= n / p; i++) {
    for (int j = 0; j <= n / q; j++) {
      if (i * p + j * q == n) {
        x = i;
        y = j;
        break;
      }
    }
  }
  if (x == -1 && y == -1)
    cout << "-1\n";
  else {
    cout << x + y << "\n";
    int ind = 0;
    for (int i = 0; i < x; i++) {
      cout << s.substr(ind, p) << "\n";
      ind += p;
    }
    for (int i = 0; i < y; i++) {
      cout << s.substr(ind, q) << "\n";
      ind += q;
    }
  }
  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>
const int MOD = 1e9 + 7;
const int N = 1e5 + 3;
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void set_bit(int& n, int b) { n |= (1ll << (b)); }
inline void unset_bit(int& n, int b) { n &= ~(1ll << (b)); }
int main() {
  int n, p, q;
  int x, y;
  string s;
  cin >> n >> p >> q;
  cin >> s;
  x = y = -1;
  for (int i = 0; i <= n / p; i++) {
    for (int j = 0; j <= n / q; j++) {
      if (i * p + j * q == n) {
        x = i;
        y = j;
        break;
      }
    }
  }
  if (x == -1 && y == -1)
    cout << "-1\n";
  else {
    cout << x + y << "\n";
    int ind = 0;
    for (int i = 0; i < x; i++) {
      cout << s.substr(ind, p) << "\n";
      ind += p;
    }
    for (int i = 0; i < y; i++) {
      cout << s.substr(ind, q) << "\n";
      ind += q;
    }
  }
  return 0;
}

```