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;
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T> p) {
  out << "(" << p.first << "," << p.second << ")";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (auto a : v) out << a << " ";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const set<T>& S) {
  for (auto a : S) cout << a << " ";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const multiset<T>& S) {
  for (auto a : S) cout << a << " ";
  return out;
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const map<S, T>& M) {
  for (auto m : M) cout << "(" << m.first << "->" << m.second << ") ";
  return out;
}
template <typename S, typename T>
pair<S, T> operator+(pair<S, T> a, pair<S, T> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename S, typename T>
pair<S, T> operator-(pair<S, T> a, pair<S, T> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
vector<int> solve(string& str, int n, int x, int y) {
  vector<int> A;
  for (int i = 0; i <= n / x; i++) {
    if ((n - i * x) % y == 0) {
      for (int j = 0; j < i; j++) {
        A.push_back(x);
      }
      for (int j = 0; j < (n - i * x) / y; j++) {
        A.push_back(y);
      }
      return A;
    }
  }
  return A;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, x, y;
  string str;
  cin >> n >> x >> y;
  cin >> str;
  vector<int> A = solve(str, n, x, y);
  if (A.empty()) {
    cout << -1 << endl;
  } else {
    int cur = 0;
    cout << A.size() << endl;
    for (int i = 0; i < A.size(); i++) {
      cout << str.substr(cur, A[i]) << '\n';
      cur += A[i];
    }
  }
  return 0;
}
