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;
void solve(const string& S, int N, int P, int Q) {
  int x, y;
  bool found = false;
  for (x = 0; x * P <= N; ++x) {
    y = (N - x * P) / Q;
    if (x * P + y * Q == N) {
      found = true;
      break;
    }
  }
  if (not found) {
    cout << -1 << endl;
    return;
  }
  cout << x + y << endl;
  int pos = 0;
  while (x--) {
    auto s = S.substr(pos, P);
    cout << s << endl;
    pos += P;
  }
  while (y--) {
    auto s = S.substr(pos, Q);
    cout << s << endl;
    pos += Q;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int N, P, Q;
  cin >> N >> P >> Q;
  string S;
  cin >> S;
  solve(S, N, P, Q);
  return 0;
}
