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 gcd(int a, int b) {
  if (a < b) return gcd(b, a);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int main(void) {
  int n, p, q;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  if (p == 1 || q == 1) {
    cout << n << endl;
    for (int i = 0; i < s.size(); i++) cout << s[i] << endl;
  } else {
    int result = 0;
    int temp = n;
    while (temp % q != 0 && temp > 0) {
      temp -= p;
      result++;
    }
    if (temp < 0) {
      cout << -1 << endl;
      return 0;
    }
    cout << result + temp / q << endl;
    for (int i = 0; i < result; i++) {
      for (int j = 0; j < p; j++) cout << s[i * p + j];
      cout << endl;
    }
    for (int i = 0; i < temp / q; i++) {
      for (int j = 0; j < q; j++) cout << s[result * p + i * q + j];
      cout << endl;
    }
  }
  return 0;
}
