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;
  int i, j, k, k1, k2, l, m;
  char s[105];
  while (scanf("%d%d%d", &n, &p, &q) == 3) {
    k = 0;
    scanf("%s", s);
    l = n / p;
    m = n / q;
    for (i = 0; i <= l; i++)
      for (j = 0; j <= m; j++) {
        if (i * p + j * q == n) {
          k = 1;
          k1 = i;
          k2 = j;
        }
      }
    if (k == 0) printf("-1\n");
    if (k == 1) {
      printf("%d\n", k1 + k2);
      for (i = 0; i < k1; i++) {
        for (j = i * p; j < i * p + p; j++) printf("%c", s[j]);
        printf("\n");
      }
      for (i = 0; i < k2; i++) {
        for (j = k1 * p + i * q; j < k1 * p + i * q + q; j++)
          printf("%c", s[j]);
        printf("\n");
      }
    }
  }
  return 0;
}
