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 long long MAX_N = 1e5 + 10;
const long long INF = 1e9 + 10;
const long long mod = 1e18;
ofstream out("output.txt");
ifstream in("intput.txt");
long long bpow(long long t, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * t) % mod;
    t = (t * t) % mod, n >>= 1;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(12);
  string s;
  long long n, p, q;
  cin >> n >> p >> q >> s;
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= n; j++) {
      cerr << i << " " << j << endl;
      if (i * p + j * q == n) {
        long long k = 0;
        cout << i + j << endl;
        for (; k < n && i; k++) {
          cout << s[k];
          if ((k + 1) % p == 0 && i) cout << endl, i--;
        }
        for (long long t = 0, aa = n - k; t < aa && j; k++, t++) {
          cout << s[k];
          if ((t + 1) % q == 0 && j) cout << endl, j--;
        }
        return 0;
      }
    }
  cout << -1;
}
