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;
int dp[114];
int last[114];
char sir[114];
vector<string> sol;
int main() {
  ios ::sync_with_stdio(false);
  int n, p, q;
  cin >> n >> p >> q;
  dp[0] = 1;
  for (int i = min(p, q); i <= n; ++i) {
    if (dp[i - p] and i - p >= 0) {
      dp[i] = 1;
      last[i] = i - p;
    } else {
      if (dp[i - q] and i - q >= 0) {
        dp[i] = 1;
        last[i] = i - q;
      }
    }
  }
  cin >> (sir + 1);
  if (dp[n] == 0) {
    cout << -1;
    exit(0);
  } else {
    while (n) {
      int next = last[n];
      string muie = "";
      while (n > next) {
        muie += sir[n];
        n--;
      }
      reverse(muie.begin(), muie.end());
      sol.push_back(muie);
    }
  }
  cout << sol.size() << '\n';
  reverse(sol.begin(), sol.end());
  for (auto x : sol) cout << x << '\n';
  return 0;
}
