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;
string str;
int n, p, q;
bool imposs;
void read() {
  cin >> n >> p >> q;
  cin >> str;
}
void solve_print() {
  int div1 = n / p, div2 = -1;
  for (; div1 >= 0; --div1) {
    if ((n - div1 * p) % q == 0) {
      div2 = (n - div1 * p) / q;
      break;
    }
  }
  if (div2 == -1) {
    cout << "-1\n";
    imposs = true;
    return;
  }
  int current = 0;
  cout << div1 + div2 << "\n";
  for (; div1 > 0; --div1) {
    int i;
    for (i = 0; i < p; ++i) {
      cout << str[current + i];
    }
    current += i;
    cout << "\n";
  }
  for (; div2 > 0; --div2) {
    int i;
    for (i = 0; i < q; ++i) {
      cout << str[current + i];
    }
    current += i;
    cout << "\n";
  }
}
int main() {
  read();
  solve_print();
  return 0;
}
