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;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
const ll MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
void setIO() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
}
int main() {
  setIO();
  int n, p, q;
  cin >> n >> p >> q;
  str second;
  cin >> second;
  for (int i = 0; i * p <= n; i++) {
    if ((n - i * p) % q == 0) {
      cout << i + (n - i * p) / q << endl;
      for (int j = 0; j < (i); ++j) cout << second.substr(j * p, p) << endl;
      for (int j = 0; j < ((n - i * p) / q); ++j)
        cout << second.substr(i * p + j * q, q) << endl;
      return 0;
    }
  }
  cout << -1 << endl;
  return 0;
}
