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 <class T>
void DBGpair(T a, string s = "") {
  return;
}
template <class T>
void DBGvec(vector<T> &a) {}
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
char dir[] = {'R', 'L', 'D', 'U'};
const int MOD = 1e9 + 7;
const long long INF = 1e9;
void setIO(string filename) {
  if (fopen((filename + ".in").c_str(), "r")) {
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
  }
}
void solve() {
  int n, p, q;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  for (int i = 0; i <= 100; i++) {
    for (int j = 0; j <= 100; j++) {
      if (p * i + q * j == n) {
        cout << i + j << "\n";
        for (int x = 0; x < p * i; x += p) cout << s.substr(x, p) << "\n";
        for (int x = p * i; x < n; x += q) cout << s.substr(x, q) << "\n";
        return;
      }
    }
  }
  cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) solve();
  return 0;
}
