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>
long long mod = 1e9 + 9;
using namespace std;
string ss;
int n;
int main() {
  int t, i, j, q, k, l, rr, n, m;
  long long su;
  string ss;
  t = 1;
  int len;
  while (t--) {
    cin >> len >> n >> m;
    cin >> ss;
    int a = -1, b = -1;
    for (i = 0; i <= 100; ++i) {
      for (j = 0; j <= 100; ++j) {
        int sum = i * n + j * m;
        if (sum == len) {
          a = i;
          b = j;
        }
      }
      if (a != -1) break;
    }
    if (a == -1)
      cout << "-1\n";
    else {
      int st = 0;
      cout << a + b << endl;
      for (i = 1; i <= a; ++i) {
        for (j = st; j <= st + n - 1; ++j) cout << ss[j];
        st += n;
        cout << endl;
      }
      for (i = 1; i <= b; ++i) {
        for (j = st; j <= st + m - 1; ++j) cout << ss[j];
        st += m;
        cout << endl;
      }
    }
  }
}
