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;
string S;
void write(int a, int b, int k, int l) {
  cout << k + l << endl;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < a; ++j) cout << S[i * a + j];
    cout << endl;
  }
  for (int i = 0; i < l; ++i) {
    for (int j = 0; j < b; ++j) cout << S[a * k + i * b + j];
    cout << endl;
  }
}
int check(int a, int b, int n) {
  int k = 1;
  if (n % a == 0) {
    write(a, b, n / a, 0);
    return 1;
  }
  if (n % b == 0) {
    write(a, b, 0, n / b);
    return 1;
  }
  while (a * k < n) {
    if ((n - a * k) % b == 0) {
      write(a, b, k, (n - a * k) / b);
      return 1;
    }
    k++;
  }
  return 0;
}
int main() {
  int n, p, q;
  cin >> n >> p >> q;
  cin >> S;
  if (check(p, q, n) == 0) cout << -1 << endl;
}
