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;
const int N = 1e3 + 5, OO = 0x3f3f3f3f, MOD = 1000000007;
const double EPS = 0.000000001;
long long sum;
long long n, m, c = 0, arr[2 * 10000 + 10], mn, mx;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int p, q, n;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  int o = n / p;
  bool t = 0, b = 0;
  int i = o, l = 0, k = 0;
  while (l <= n && i >= 0) {
    if (n - p * i - q * l == 0) {
      t = 1;
      break;
    } else if (n - p * i - q * l > 0) {
      l++;
    } else if (n - p * i - q * l < 0) {
      i--;
    }
  }
  if (!t) {
    cout << -1;
  } else {
    cout << l + i << endl;
    for (int w = 0; w < i; w++) {
      for (int j = 0; j < p; j++) {
        cout << s[k], k++;
      }
      cout << endl;
    }
    for (int w = 0; w < l; w++) {
      for (int j = 0; j < q; j++) {
        cout << s[k], k++;
      }
      cout << endl;
    }
  }
}
