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() {
  char texto[100], ch;
  int i, j, k, cont, p, q, x, y, aux, tam, qnt;
  int pares[tam];
  scanf("%d %d %d ", &tam, &p, &q);
  for (i = 0; i < tam; i++) {
    scanf("%c", &texto[i]);
  }
  k = 0;
  for (i = 0; i <= tam; i++) {
    for (j = 0; j <= tam; j++) {
      if (p * i + q * j == tam) {
        x = i;
        y = j;
        k = 1;
      }
    }
  }
  if (k == 0) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", x + y);
  j = 0;
  while (x > 0) {
    for (i = 0; i < p; i++) {
      printf("%c", texto[j]);
      j++;
    }
    printf("\n");
    x--;
  }
  while (y > 0) {
    for (i = 0; i < q; i++) {
      printf("%c", texto[j]);
      j++;
    }
    printf("\n");
    y--;
  }
  return 0;
}
