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(0);
  cin.tie(0);
  cout.tie(0);
  int n, p, q, x, y, pos = 0;
  string s;
  bool found = false;
  cin >> n >> p >> q >> s;
  for (x = 0; x <= n; x++) {
    for (y = 0; y <= n; y++) {
      if (p * x + q * y == n) {
        found = true;
        break;
      }
    }
    if (found) break;
  }
  if (found) {
    cout << x + y << endl;
    while (x) {
      cout << s.substr(pos, p) << endl;
      pos += p;
      x--;
    }
    while (y) {
      cout << s.substr(pos, q) << endl;
      pos += q;
      y--;
    }
  } else
    cout << -1;
  return 0;
}
