#include <bits/stdc++.h>
using namespace std;
int n, k, a, b, fst, sec;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> a >> b >> s;
  if (a > b) swap(a, b);
  k = n;
  while (k % b != 0 && k > 0) {
    k -= a;
    ++fst;
  }
  if (k % b == 0) sec = k / b, k = 0;
  if (k != 0)
    cout << -1 << "\n";
  else {
    int last = 0;
    cout << fst + sec << "\n";
    for (int i = 0; i < fst; i++) {
      for (int j = last; j < a + last; j++) cout << s[j];
      cout << "\n";
      last += a;
    }
    for (int i = 0; i < sec; i++) {
      for (int j = last; j < b + last; j++) cout << s[j];
      cout << "\n";
      last += b;
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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 n, k, a, b, fst, sec;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> a >> b >> s;
  if (a > b) swap(a, b);
  k = n;
  while (k % b != 0 && k > 0) {
    k -= a;
    ++fst;
  }
  if (k % b == 0) sec = k / b, k = 0;
  if (k != 0)
    cout << -1 << "\n";
  else {
    int last = 0;
    cout << fst + sec << "\n";
    for (int i = 0; i < fst; i++) {
      for (int j = last; j < a + last; j++) cout << s[j];
      cout << "\n";
      last += a;
    }
    for (int i = 0; i < sec; i++) {
      for (int j = last; j < b + last; j++) cout << s[j];
      cout << "\n";
      last += b;
    }
  }
  return 0;
}

```