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
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int a, b, c;
  int p = -1, q = -1, x = -1, y = -1, z = 0;
  string sample;
  cin >> a >> b >> c >> sample;
  if (b == c) {
    if (a % b == 0) {
      cout << a / b << endl;
      for (int i = 0; i < a; i += b) {
        cout << sample.substr(i, b) << endl;
      }
    } else {
      cout << "-1\n";
      return 0;
    }
  } else if (a % b == 0) {
    cout << a / b << endl;
    for (int i = 0; i < a; i += b) {
      cout << sample.substr(i, b) << endl;
    }
  } else if (a % c == 0) {
    cout << a / c << endl;
    for (int i = 0; i < a; i += c) {
      cout << sample.substr(i, c) << endl;
    }
  } else {
    if (b > c) {
      q = b;
      p = c;
    } else {
      q = c, p = b;
    }
    for (int i = 1; i <= a / p; i++) {
      if ((a - i * p) % q == 0) {
        x = i;
        y = (a - i * p) / q;
        z = 1;
        break;
      }
    }
    if (z) {
      cout << x + y << endl;
      for (int i = 0; i < p * x; i += p) {
        cout << sample.substr(i, p) << endl;
      }
      for (int i = p * x; i < a; i += q) {
        cout << sample.substr(i, q) << endl;
      }
    } else {
      cout << "-1\n";
    }
    return 0;
  }
  return 0;
}
