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;
stack<string> s;
int dp[1000];
bool fn(string s1, int n, int p, int q, int curr) {
  if (curr == n) {
    dp[curr] = 1;
    return true;
  } else if (curr > n) {
    dp[curr] = -1;
    return false;
  }
  bool a;
  if (dp[curr + p] == 1)
    a = true;
  else if (dp[curr + p] == -1)
    a = false;
  else
    a = fn(s1, n, p, q, curr + p);
  if (a) {
    s.push(s1.substr(curr, p));
    dp[curr] = 1;
    return true;
  }
  if (dp[curr + q] == 1)
    a = true;
  else if (dp[curr + q] == -1)
    a = false;
  a = fn(s1, n, p, q, curr + q);
  if (a) {
    s.push(s1.substr(curr, q));
    dp[curr] = 1;
    return true;
  }
  dp[curr] = -1;
  return false;
}
int main() {
  int n, p, q;
  cin >> n >> p >> q;
  string s1;
  cin >> s1;
  bool a = fn(s1, n, p, q, 0);
  if (!a)
    cout << -1;
  else {
    cout << s.size() << endl;
    while (!s.empty()) {
      cout << s.top() << endl;
      s.pop();
    }
  }
  return 0;
}
