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, c = 0, d;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  if (n % p == 0) {
    d = n / p;
    cout << d;
    for (i = 0; i < n; i++, c++) {
      if (c % p == 0) cout << endl;
      cout << s[i];
    }
  } else if (n % q == 0) {
    d = n / q;
    cout << d;
    for (i = 0; i < n; i++, c++) {
      if (c % q == 0) cout << endl;
      cout << s[i];
    }
  } else {
    int b = 1, flag = 0;
    int x = b * q, u;
    while (n >= x) {
      u = n - x;
      if (u % p == 0) {
        flag = 1;
        break;
      }
      b++;
      x = b * q;
    }
    if (flag) {
      int a = u / p;
      cout << a + b;
      for (i = 0; i < p * a; i++, c++) {
        if (c % p == 0) cout << endl;
        cout << s[i];
      }
      c = 0;
      for (int j = i; j < n; j++, c++) {
        if (c % q == 0) cout << endl;
        cout << s[j];
      }
    } else
      cout << -1;
  }
  return 0;
}
