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;
long long second, t, n, m, i, j, l, r, k, p, q, ans, mx, my, first;
string a, b;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> l >> r >> a;
  m = -1, k = -1;
  if (n % l == 0)
    m = n / l, k = 0;
  else if (n % r == 0)
    m = 0, k = n / r;
  for (i = 1; i <= n; i++)
    if ((n - i * l) % r == 0 && n >= i * l) {
      m = i, k = (n - i * l) / r;
      break;
    }
  if (m == -1 || k == -1) {
    cout << -1;
    return 0;
  }
  n = 0;
  cout << m + k << '\n';
  for (i = 1; i <= m; i++) {
    b = "";
    for (j = 0; j < l; j++, n++) b += a[n];
    cout << b << '\n';
  }
  for (i = 1; i <= k; i++) {
    b = "";
    for (j = 0; j < r; j++, n++) b += a[n];
    cout << b << '\n';
  }
}
