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;
void run() {
  int n, p, q;
  string s;
  cin >> n >> p >> q;
  cin >> s;
  int dem = 0;
  int x = 0, y = 0;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      if (s.size() == i * p + j * q) {
        x = i;
        y = j;
        dem++;
      }
  if (x + y != 0) cout << x + y << endl;
  if (x != 0 || y != 0) {
    int kt = 0, d = 0;
    int i = 0, j;
    if (x != 0) {
      for (i = 0; i < s.size(); i++) {
        if (d == p) {
          cout << endl;
          i--;
          kt++;
          d = 0;
        } else {
          cout << s[i];
          d++;
        }
        if (kt == x) break;
      }
      kt = 0, d = 0;
      for (j = i + 1; j < s.size(); j++) {
        if (d == q) {
          cout << endl;
          j--;
          kt++;
          d = 0;
        } else {
          cout << s[j];
          d++;
        }
        if (kt == y) {
          break;
        }
      }
    } else {
      kt = 0, d = 0;
      for (j = 0; j < s.size(); j++) {
        if (d == q) {
          cout << endl;
          j--;
          kt++;
          d = 0;
        } else {
          cout << s[j];
          d++;
        }
        if (kt == y) {
          break;
        }
      }
    }
  } else {
    cout << "-1";
  }
}
int main() {
  run();
  return 0;
}
