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, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.s << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  int l = v.size();
  for (int i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int inf = 1e9 + 10;
const long long int INF = 1e18 + 10;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 10;
void solve() {
  int n, p, q;
  string s;
  cin >> n >> p >> q;
  cin >> s;
  int x, y;
  bool boo = 0;
  for (x = 0; x <= n; x++) {
    for (y = 0; y <= n; y++) {
      if (x * p + y * q == n) {
        boo = 1;
        break;
      }
    }
    if (boo) break;
  }
  if (!boo) {
    cout << -1 << endl;
    return;
  }
  vector<string> s1;
  int i = 0;
  while (i < x * p) {
    s1.push_back(s.substr(i, p));
    i += p;
  }
  while (i < n) {
    s1.push_back(s.substr(i, q));
    i += q;
  }
  cout << s1.size() << endl;
  for (int i = 0; i < s1.size(); i += 1) {
    cout << s1[i] << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
