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 INF = int(1e9);
const int MOD = int(1e9) + 7;
const long long N = 1000010;
const long long INFll = 1ll * INF * INF;
const long double ldINF = 1e+018;
const long double EPS = 0.000000001;
const int sdx[4] = {-1, 0, 1, 0};
const int sdy[4] = {0, 1, 0, -1};
const long double PI = 3.1415926535897932384;
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, pair<T, T>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i] << endl;
    else
      out << a[i] << endl;
  return out;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, p, q;
  cin >> n >> p >> q;
  vector<char> c(n + 1);
  for (int i = 1; i <= n; i++) cin >> c[i];
  vector<int> dp(1000);
  dp[0] = 1;
  for (int x = 0; x <= n; x++)
    if (dp[x]) dp[x + p] = 1;
  for (int x = 0; x <= n; x++)
    if (dp[x]) dp[x + q] = 1;
  if (dp[n] == 0) {
    cout << -1 << endl;
    return 0;
  }
  vector<string> ans;
  int x = n;
  while (x > 0) {
    if (x - p >= 0 && dp[x - p]) {
      string s = "";
      for (int i = x - p + 1; i <= x; i++) s += c[i];
      ans.push_back(s);
      x -= p;
    } else if (x - q >= 0 && dp[x - q]) {
      string s = "";
      for (int i = x - q + 1; i <= x; i++) s += c[i];
      ans.push_back(s);
      x -= q;
    }
  }
  reverse(ans.begin(), ans.end());
  cout << ans.size() << endl;
  cout << ans << endl;
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}
