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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int sz, a, b, x = 0, y = 0;
  string str;
  cin >> sz >> a >> b;
  cin >> str;
  for (int i = 0; i <= 100; i++) {
    int j = (sz - a * i) / b;
    if (j < 0) break;
    if (j > 0 && a * i + b * j == sz) {
      x = i;
      y = j;
      break;
    }
  }
  if (x == 0 && y == 0) {
    for (int i = 0; i <= 100; i++) {
      int j = (sz - b * i) / a;
      if (j < 0) break;
      if (j > 0 && b * i + a * j == sz) {
        x = j;
        y = i;
        break;
      }
    }
  }
  if (x == 0 && y == 0)
    cout << -1 << "\n";
  else {
    cout << x + y << "\n";
    for (int i = 0; i < x * a; i += a) {
      cout << str.substr(i, a) << "\n";
    }
    for (int j = x * a; j < sz; j += b) {
      cout << str.substr(j, b) << "\n";
    }
  }
}
