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

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

```