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;
using LL = int64_t;
using pii = pair<int, int>;
const double PI = 2 * asin(1);
int g_n, g_p, g_q;
string g_s;
void pretreat() {}
void input() { cin >> g_s; }
void solve() {
  bool f = false;
  int p, q;
  string s1, s2;
  for (int i = 0; i <= g_n / g_p; ++i) {
    for (int j = 0; j <= g_n / g_q; ++j) {
      if (i * g_p + j * g_q != g_n) continue;
      p = i, q = j, f = true;
      break;
    }
  }
  if (!f) {
    puts("-1");
    return;
  }
  cout << p + q << endl;
  for (int i = 0; i < p; ++i) {
    cout << string(g_s.begin() + i * g_p, g_s.begin() + (i + 1) * g_p) << endl;
  }
  for (int i = 0; i < q; ++i) {
    cout << string(g_s.begin() + p * g_p + i * g_q,
                   g_s.begin() + p * g_p + (i + 1) * g_q)
         << endl;
  }
}
int main() {
  pretreat();
  while (cin >> g_n >> g_p >> g_q) {
    input();
    solve();
  }
  return 0;
}
