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 long long n, p, q;
string s;
bool used[1005];
vector<int> my;
void rec(int sum) {
  if (sum > n || used[sum]) {
    my.pop_back();
    return;
  }
  used[sum] = 1;
  if (sum == n) {
    int k = 0;
    string s1 = "";
    cout << my.size() << endl;
    for (int i = 0; i < my.size(); i++) {
      s1 = "";
      for (int j = k; j < k + my[i]; j++) {
        s1 += s[j];
      }
      k += my[i];
      cout << s1 << endl;
    }
    exit(0);
  }
  my.push_back(p);
  rec(sum + p);
  my.push_back(q);
  rec(sum + q);
  my.pop_back();
}
int main() {
  cin >> n >> p >> q;
  cin >> s;
  rec(0);
  cout << "-1";
}
