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>
#pragma GCC optimize("O3")
using namespace std;
const long long INF = 2e9;
const long long MOD = 1e9 + 7;
const long long MB = 20;
void solve() {
  long long n, p, q;
  string s;
  cin >> n >> p >> q >> s;
  for (long long i = 0; p * i <= n; i++) {
    if ((n - p * i) % q) continue;
    long long ans = i + (n - p * i) / q;
    cout << ans << '\n';
    for (long long j = 0, c = 0; c < i; j += p, c++) {
      cout << s.substr(j, p) << '\n';
    }
    for (long long j = p * i; j < n; j += q) {
      cout << s.substr(j, q) << '\n';
    }
    return;
  }
  cout << -1;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(12);
  solve();
}
