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 n, p, q, a[105];
string s;
pair<int, int> sol[101];
int main() {
  cin >> n >> p >> q >> s;
  a[0] = true;
  for (int j = 0; j <= max(0, n - p); j++)
    if (a[j + p] == false && a[j] == true) {
      a[j + p] = true;
      sol[j + p] = make_pair(sol[j].first + 1, sol[j].second);
    }
  for (int j = 0; j <= max(0, n - q); j++)
    if (a[j + q] == false && a[j] == true) {
      a[j + q] = true;
      sol[j + q] = make_pair(sol[j].first, sol[j].second + 1);
    }
  if (a[n] == false)
    cout << "-1\n";
  else {
    cout << sol[n].first + sol[n].second << '\n';
    int k = 0;
    for (int i = 1; i <= sol[n].first; i++) {
      for (int j = 1; j <= p; j++) {
        cout << s[k];
        k++;
      }
      cout << '\n';
    }
    for (int i = 1; i <= sol[n].second; i++) {
      for (int j = 1; j <= q; j++) {
        cout << s[k];
        k++;
      }
      cout << '\n';
    }
  }
  return 0;
}
