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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, p, q;
  string a;
  cin >> n >> p >> q >> a;
  if (p > q) {
    swap(p, q);
  }
  if (p > n && q > n) {
    cout << -1 << "\n";
    return 0;
  }
  if (n % p == 0) {
    cout << n / p << "\n";
    for (int i = 1; i <= n; ++i) {
      cout << a[i - 1];
      if (i % p == 0) cout << "\n";
    }
    return 0;
  } else if (n % q == 0) {
    cout << n / q << "\n";
    for (int i = 1; i <= n; ++i) {
      cout << a[i - 1];
      if (i % q == 0) cout << "\n";
    }
    return 0;
  }
  int res1 = 0;
  bool ok = 1;
  for (int i = 0; i <= n / p; ++i) {
    if (p * i + (n - p * i) / q * q == n) {
      ok = 0;
      res1 = i;
    }
  }
  if (ok) {
    cout << -1 << "\n";
    return 0;
  }
  cout << res1 + (n - p * res1) / q << "\n";
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    if (res1 == 0) {
      res = i;
      break;
    }
    cout << a[i - 1];
    if (i % p == 0) {
      cout << "\n";
      --res1;
    }
  }
  for (int i = res; i <= n; ++i) {
    cout << a[i - 1];
    if ((i - res + 1) % q == 0) {
      cout << "\n";
    }
  }
}
