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;
vector<int> ret;
int p, q, n;
bool dp[101];
bool sol(int ind) {
  if (ind == n) return true;
  if (ind > n) return false;
  if (dp[ind]) return false;
  dp[ind] = true;
  if (sol(ind + p)) {
    ret.push_back(p);
    return true;
  }
  if (sol(ind + q)) {
    ret.push_back(q);
    return true;
  }
  return false;
}
int main() {
  string s;
  cin >> n >> p >> q >> s;
  if (!sol(0))
    cout << -1 << endl;
  else {
    reverse(ret.begin(), ret.end());
    cout << ret.size() << endl;
    int x, pos = 0, j, i;
    for (i = 0; i < ret.size(); ++i) {
      x = ret[i];
      for (j = pos; j < pos + x; ++j) putchar(s[j]);
      putchar('\n');
      pos += x;
    }
  }
}
