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() {
  int n, p, q, count = 1, buf = -1, ctr = 1, col = 0;
  cin >> n >> p >> q;
  string str;
  cin >> str;
  int max, min;
  if (p > q) {
    max = p;
    min = q;
  } else {
    max = q;
    min = p;
  }
  int ost = n - max;
  while ((ost >= min) || (n % p == 0) && buf == -1 ||
         (n % q == 0) && buf == -1) {
    if (n % p == 0) {
      cout << n / p;
      for (int i = 0; i < n; ++i) {
        if (i % p == 0) cout << endl;
        cout << str[i];
      }
      ost = 0;
      buf++;
    } else if (n % q == 0) {
      cout << n / q;
      for (int i = 0; i < n; ++i) {
        if (i % q == 0) cout << endl;
        cout << str[i];
      }
      ost = 0;
      buf++;
    } else if (ost % min == 0) {
      cout << count + ost / min << endl;
      for (int i = 0; i < n; ++i) {
        cout << str[i];
        if ((i < max * count) && ((i + 1) % max == 0) && (i != 0) &&
            (n - col > ost))
          cout << endl;
        if (((n - ctr) % min == 0) && (n - col <= ost)) {
          cout << endl;
          ctr++;
        } else
          ctr++;
        col++;
      }
      ost = 0;
      buf++;
    } else if (ost > min) {
      ost -= max;
      count++;
    }
  }
  if (buf == -1) cout << -1;
}
