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;
inline void boost() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
}
const long long maxn = 1e5 + 123;
const long long inf = 1e9 + 123;
const long long mod = 1e9 + 7;
const double pi = acos(-1);
int main() {
  boost();
  int n, p, q;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  if (n % p == 0) {
    cout << n / p << "\n";
    int it = 0;
    while (it <= s.size() - 1) {
      cout << s.substr(it, p) << "\n";
      it += p;
    }
    return 0;
  }
  if (n % q == 0) {
    cout << n / q << "\n";
    int it = 0;
    while (it <= s.size() - 1) {
      cout << s.substr(it, q) << "\n";
      it += q;
    }
    return 0;
  }
  int cntr = 0;
  int kol = 0;
  while (cntr < n) {
    cntr += p;
    kol++;
    if ((n - cntr) % q == 0 && cntr < n) {
      break;
    }
  }
  if ((n - cntr) % q == 0 && cntr < n) {
    int it = 0;
    cout << kol + (n - cntr) / q << "\n";
    while (kol--) {
      cout << s.substr(it, p) << "\n";
      it += p;
    }
    while (it <= s.size() - 1) {
      cout << s.substr(it, q) << "\n";
      it += q;
    }
    return 0;
  }
  kol = 0;
  cntr = 0;
  while (cntr < n) {
    cntr += q;
    kol++;
    if ((n - cntr) % p == 0 && cntr < n) {
      break;
    }
  }
  if ((n - cntr) % p == 0 && cntr < n) {
    int it = 0;
    cout << kol + (n - cntr) / p << "\n";
    while (kol--) {
      cout << s.substr(it, q) << "\n";
      it += q;
    }
    while (it <= s.size() - 1) {
      cout << s.substr(it, p) << "\n";
      it += p;
    }
    return 0;
  }
  cout << "-1";
  return 0;
}
