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;
inline long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  string s;
  int n, a, b;
  cin >> n >> a >> b >> s;
  for (int i = 0; i <= n / a; i++) {
    if ((n - i * a) % b == 0) {
      cout << i + (n - i * a) / b << endl;
      for (int j = 0; j < i * a; j++) {
        cout << s[j];
        if ((j + 1) % a == 0) cout << endl;
      }
      for (int j = i * a, k = 0; j < n; j++, k++) {
        cout << s[j];
        if ((k + 1) % b == 0) cout << endl;
      }
      return 0;
    }
  }
  cout << "-1\n";
  return 0;
}
