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, i = 1, j, k, temp = 0, flag = 0;
  string s;
  cin >> n >> p >> q;
  cin >> s;
  if (n % p == 0 && n != p + q || n % q == 0 && n != p + q) {
    if (n % p == 0) {
      cout << n / p << endl;
      for (i = 0; i < n; i++) {
        cout << s[i];
        if ((i + 1) % p == 0) cout << endl;
      }
    } else {
      cout << n / q << endl;
      for (i = 0; i < n; i++) {
        cout << s[i];
        if ((i + 1) % q == 0) cout << endl;
      }
    }
  } else {
    while (1) {
      j = n - (p * i);
      k = n - (q * i);
      if (j < 0 || k < 0) break;
      if (j % q == 0) {
        temp = 1;
        break;
      }
      if (k % p == 0) {
        flag = 1;
        break;
      }
      if (j < 0 && k < 0) break;
      i++;
    }
    if (temp == 1) {
      cout << i + (j / q) << endl;
      for (int x = 0; x < i * p; x++) {
        cout << s[x];
        if ((x + 1) % p == 0) cout << endl;
      }
      j = 1;
      for (int x = i * p; x < n; x++) {
        cout << s[x];
        if ((j) % q == 0) cout << endl;
        j++;
      }
    } else if (flag == 1) {
      cout << i + (k / p) << endl;
      for (int x = 0; x < i * q; x++) {
        cout << s[x];
        if ((x + 1) % q == 0) cout << endl;
      }
      j = 1;
      for (int x = i * q; x < n; x++) {
        cout << s[x];
        if ((j) % p == 0) cout << endl;
        j++;
      }
    } else {
      cout << "-1" << endl;
    }
  }
}
