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>
int main() {
  int n, p, q, i, j, flag = 0, ans, k, a;
  scanf("%d%d%d", &n, &p, &q);
  char s[n];
  scanf("%s", s);
  for (i = 0; p * i <= n; i++) {
    for (j = 0; (p * i + q * j) <= n; j++) {
      ans = p * i + q * j;
      if (ans == n) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  if (flag == 1) {
    k = 0;
    printf("%d\n", i + j);
    while (i--) {
      a = p;
      while (a--) {
        printf("%c", s[k]);
        k++;
      }
      printf("\n");
    }
    while (j--) {
      a = q;
      while (a--) {
        printf("%c", s[k]);
        k++;
      }
      printf("\n");
    }
  } else
    printf("-1\n");
  return 0;
}
