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>
char s[105];
int main() {
  int n, p, q;
  scanf("%d%d%d", &n, &p, &q);
  scanf("%s", s);
  if (n < p && n < q) {
    puts("-1");
    return 0;
  }
  if (n % p == 0) {
    printf("%d\n", n / p);
    for (int i = 0; i < n; i++) {
      if (i % p == 0 && i != 0) printf("\n");
      printf("%c", s[i]);
    }
    return 0;
  }
  if (n % q == 0) {
    printf("%d\n", n / q);
    for (int i = 0; i < n; i++) {
      if (i % q == 0 && i != 0) printf("\n");
      printf("%c", s[i]);
    }
    return 0;
  }
  int pos = 0;
  for (; pos < n + 1; pos += p)
    if ((n - pos) % q == 0) break;
  if (pos >= n)
    puts("-1");
  else {
    int j = 0;
    printf("%d\n", pos / p + (n - pos) / q);
    for (int i = 0; i < pos / p; i++) {
      for (int k = 0; k < p; k++, j++) printf("%c", s[j]);
      printf("\n");
    }
    for (int i = 0; i < (n - pos) / q; i++) {
      for (int k = 0; k < q; k++, j++) printf("%c", s[j]);
      printf("\n");
    }
  }
  return 0;
}
