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() {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  int x, y, n;
  cin >> n >> x >> y;
  string s;
  cin >> s;
  vector<string> vec;
  int ch = -1, a = 0, b = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (i * x + j * y == n) ch = i, a = i, b = j;
    }
  }
  if (ch == -1)
    cout << -1 << endl;
  else if (n % y == 0) {
    for (int i = 0; i < n; i += y) vec.push_back(s.substr(i, y));
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << endl;
    return 0;
  } else if (n % x == 0) {
    for (int i = 0; i < n; i += x) vec.push_back(s.substr(i, x));
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << endl;
    return 0;
  } else if (n % x == y) {
    vec.push_back(s.substr(0, y));
    for (int i = y; i < n; i += x) vec.push_back(s.substr(i, x));
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << endl;
    return 0;
  } else if (n % y == x) {
    vec.push_back(s.substr(0, x));
    for (int i = x; i < n; i += y) vec.push_back(s.substr(i, y));
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << endl;
    return 0;
  } else {
    int c = 0;
    while (a--) {
      vec.push_back(s.substr(c, x));
      c += x;
    }
    while (b--) vec.push_back(s.substr(c, y)), c += y;
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << endl;
  }
  return 0;
}
