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() {
  int n, p, q;
  scanf("%d %d %d", &n, &p, &q);
  string s;
  cin >> s;
  if (n < q && n < p) {
    cout << -1 << endl;
    return 0;
  } else {
    for (int i = 0; i <= n / p; ++i) {
      for (int j = 0; j <= n / q; ++j) {
        if ((i * p) + (j * q) == n) {
          cout << i + j << '\n';
          for (int a = 0; a < i; ++a) {
            cout << s.substr(a * p, p) << '\n';
          }
          for (int b = 0; b < j; ++b) {
            cout << s.substr((i * p) + (b * q), q) << '\n';
          }
          return 0;
        }
      }
    }
  }
  cout << -1 << endl;
  return 0;
}
