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;
const int maxn = 1005;
const int inf = 1 << 28;
int main() {
  int n, p, q;
  cin >> n >> p >> q;
  string str;
  cin >> str;
  int N = n;
  if (N % p == 0) {
    cout << N / p << endl;
    for (int i = 0; i < N / p; i++) {
      cout << str.substr(i * p, p) << endl;
      ;
    }
    return 0;
  }
  if (N % q == 0) {
    cout << N / q << endl;
    for (int i = 0; i < N / q; i++) {
      cout << str.substr(i * q, q) << endl;
      ;
    }
    return 0;
  }
  int x = 1, y = 1;
  for (x = 1; x <= N; x++) {
    for (y = 1; y <= N; y++)
      if (x * p + y * q == N) {
        goto A;
      }
  }
A:;
  if (x * p + y * q == N) {
    cout << x + y << endl;
    for (int i = 0; i < x; i++) {
      cout << str.substr(i * p, p) << endl;
    }
    for (int i = 0; i < y; i++) {
      cout << str.substr(x * p + i * q, q) << endl;
    }
    return 0;
  }
  return puts("-1");
  return 0;
}
