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;
char a[200];
int main() {
  int n, p, q;
  cin >> n >> p >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (p == 1 || q == 1) {
    cout << n << endl;
    for (int i = 1; i <= n; i++) cout << a[i] << endl;
    return 0;
  } else {
    if (p + q == n) {
      cout << "2" << endl;
      for (int i = 1; i <= p; i++) cout << a[i];
      cout << endl;
      for (int i = p + 1; i <= n; i++) cout << a[i];
      cout << endl;
      return 0;
    } else {
      if (n % p == 0) {
        cout << n / p << endl;
        for (int i = 1; i <= n; i++) {
          cout << a[i];
          if (i % p == 0) cout << endl;
        }
        return 0;
      }
      if (n % q == 0) {
        cout << n / q << endl;
        for (int i = 1; i <= n; i++) {
          cout << a[i];
          if (i % q == 0) cout << endl;
        }
        return 0;
      }
      int mi = min(p, q);
      int mx = n / mi + 1;
      int x = 0;
      int y = 0;
      for (int i = 1; i <= mx; i++) {
        x = i * p;
        y = n - x;
        if (y < 0) break;
        if (y % q == 0) {
          cout << i + y / q << endl;
          for (int i = 1; i <= x; i++) {
            cout << a[i];
            if (i % p == 0) cout << endl;
          }
          for (int i = x + 1; i <= n; i++) {
            cout << a[i];
            if ((i - x) % q == 0) cout << endl;
          }
          return 0;
        }
      }
    }
  }
  cout << "-1" << endl;
  return 0;
}
