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;
const int inf = 1e9 + 7;
int x, y, z;
string str;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> z >> x >> y >> str;
  for (int a = 0; a <= 100; a++) {
    int b;
    if (((z - a * x) % y) == 0) {
      b = (z - a * x) / y;
      if (!(0 <= b && b <= 100)) continue;
      cout << a + b << '\n';
      int i = 0;
      while (a--) {
        for (int j = 1; j <= x; j++) {
          cout << str[i++];
        }
        cout << '\n';
      }
      while (b--) {
        for (int j = 1; j <= y; j++) {
          cout << str[i++];
        }
        cout << '\n';
      }
      return 0;
    }
  }
  cout << -1;
}
