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 n, a, b, res;
string ans, s;
vector<string> v;
int main() {
  cin >> n >> a >> b;
  cin >> s;
  if (n % a != 0 && n % b != 0) {
    int m = n;
    while (m % b != 0 && m > 0) m -= a;
    if (m % b == 0 && m > 0) {
      for (int i = 0; i < n - m; i++) {
        ans += s[i];
        if (ans.size() % a == 0) {
          v.push_back(ans);
          ans = "";
        }
      }
      s.erase(0, n - m);
      n = m;
    } else {
      m = n;
      while (m % a != 0 && m > 0) m -= b;
      if (m % a == 0 && m > 0) {
        for (int i = 0; i < n - m; i++) {
          ans += s[i];
          if (ans.size() % b == 0) {
            v.push_back(ans);
            ans = "";
          }
        }
        n = m;
      }
      s.erase(0, n - m);
    }
  }
  if (n % a == 0) {
    for (int i = 0; i < n; i++) {
      ans += s[i];
      if (ans.size() % a == 0) {
        v.push_back(ans);
        ans = "";
      }
    }
  } else if (n % b == 0) {
    for (int i = 0; i < n; i++) {
      ans += s[i];
      if (ans.size() % b == 0) {
        v.push_back(ans);
        ans = "";
      }
    }
  } else {
    puts("-1");
    return 0;
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << endl;
  }
}
