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 N6 = 1e6 + 6, N3 = 1e3 + 6, oo = 1e9 + 9;
const long long ool = 1e18 + 9;
int n, p, q;
string s;
vector<string> ans;
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> n >> p >> q;
  cin >> s;
  int x = 0;
  int cnt = 0;
  while ((n - x) % q != 0 && x <= n) {
    ++cnt;
    x += p;
  }
  if (x <= n && (n - x) % q == 0) {
    int cur = 0;
    for (int i = 1; i <= cnt; ++i) {
      string t = "";
      for (int j = cur; j <= cur + p - 1; ++j) {
        t += s[j];
      }
      cur += p;
      ans.push_back(t);
    }
    for (int i = 1; i <= (n - x) / q; ++i) {
      string t = "";
      for (int j = cur; j <= cur + q - 1; ++j) t += s[j];
      cur += q;
      ans.push_back(t);
    }
    cout << ans.size() << "\n";
    for (auto it : ans) cout << it << "\n";
    return 0;
  }
  cout << -1;
  return 0;
}
