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;
int main() {
  ios_base::sync_with_stdio(0);
  std::cin.tie(0);
  int n, p, q, x, y;
  bool exists = false;
  char s[105];
  scanf("%d%d%d\n", &n, &p, &q);
  scanf("%s", s);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      if ((p * i) + (q * j) == n) {
        exists = true;
        x = i;
        y = j;
      }
  if (exists) {
    printf("%d\n", x + y);
    int offset;
    for (int k = 0; k < x; k++) {
      offset = k * p;
      for (int i = offset; i < offset + p; i++) printf("%c", s[i]);
      puts("");
    }
    for (int k = 0; k < y; k++) {
      offset = (x * p) + (k * q);
      for (int i = offset; i < offset + q; i++) printf("%c", s[i]);
      puts("");
    }
  } else {
    puts("-1");
  }
  return 0;
}
