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, x, y;
  string s;
  cin >> n >> x >> y;
  cin >> s;
  if (x + y > n && n % min(x, y) == 0) {
    int splitter = min(x, y);
    cout << n / splitter << endl;
    for (int i = 0; i < n; i += splitter) {
      cout << s.substr(i, splitter) << endl;
    }
    return 0;
  }
  if (x + y > n) {
    if (max(x, y) == n)
      cout << 1 << endl << s;
    else
      cout << -1;
    return 0;
  }
  if (x + y == n) {
    cout << 2 << endl;
    cout << s.substr(0, x) << endl << s.substr(x);
    return 0;
  }
  for (int i = 0; (n % x + x * i) < n; i++) {
    if ((n % x + x * i) % y == 0) {
      int ycount = (n % x + x * i) / y;
      int xcount = (n - ycount * y) / x;
      cout << ycount + xcount << endl;
      for (int i = 0; i < xcount * x; i += x) {
        cout << s.substr(i, x) << endl;
      }
      for (int i = xcount * x; i < n; i += y) {
        cout << s.substr(i, y) << endl;
      }
      return 0;
    }
  }
  for (int i = 0; (n % y + y * i) < n; i++) {
    if ((n % y + y * i) % x == 0) {
      int xcount = (n % y + y * i) / x;
      int ycount = (n - xcount * x) / y;
      cout << ycount + xcount << endl;
      for (int i = 0; i < xcount * x; i += x) {
        cout << s.substr(i, x) << endl;
      }
      for (int i = xcount * x; i < n; i += y) {
        cout << s.substr(i, y) << endl;
      }
      return 0;
    }
  }
  cout << -1;
  return 0;
}
