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;
char str[10005];
int main() {
  ios::sync_with_stdio(false);
  int n, a, b;
  bool f = 0;
  int i, len, lena;
  cin >> n >> a >> b;
  cin >> str + 1;
  if (n % a == 0) {
    printf("%d\n", n / a);
    for (i = 1; i <= n; i++) {
      f = 1;
      printf("%c", str[i]);
      if (i % a == 0 && i < n) puts("");
    }
  } else {
    len = n;
    lena = 0;
    while (f == 0 && lena <= n) {
      if (len % b == 0) {
        printf("%d\n", lena / a + len / b);
        for (i = 1; i <= n; i++) {
          f = 1;
          printf("%c", str[i]);
          if (i % a == 0 && i < lena) puts("");
          if ((i - lena) % (b) == 0 && i >= lena) puts("");
        }
      }
      len -= a;
      lena += a;
    }
  }
  if (f == 0) puts("-1");
}
