#include <bits/stdc++.h>
int main() {
  int n, p, q, i, j, k, flag = 0, m = 0, k1;
  scanf("%d %d %d ", &n, &p, &q);
  char a[n + 1];
  scanf("%s", a);
  for (i = 0; i <= n / p; i++) {
    for (j = 0; j <= n / q; j++) {
      if (((p * i) + (q * j)) == n) {
        flag = 1;
        break;
      } else if ((p * i) + (q * j) > n)
        break;
    }
    if (flag) break;
  }
  if (flag) {
    printf("%d\n", i + j);
    for (k = 0; k < i; k++) {
      for (k1 = 0; k1 < p; k1++) {
        printf("%c", a[m]);
        m++;
      }
      printf("\n");
    }
    for (k = 0; k < j; k++) {
      for (k1 = 0; k1 < q; k1++) {
        printf("%c", a[m]);
        m++;
      }
      printf("\n");
    }
  } else
    printf("-1\n");
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int n, p, q, i, j, k, flag = 0, m = 0, k1;
  scanf("%d %d %d ", &n, &p, &q);
  char a[n + 1];
  scanf("%s", a);
  for (i = 0; i <= n / p; i++) {
    for (j = 0; j <= n / q; j++) {
      if (((p * i) + (q * j)) == n) {
        flag = 1;
        break;
      } else if ((p * i) + (q * j) > n)
        break;
    }
    if (flag) break;
  }
  if (flag) {
    printf("%d\n", i + j);
    for (k = 0; k < i; k++) {
      for (k1 = 0; k1 < p; k1++) {
        printf("%c", a[m]);
        m++;
      }
      printf("\n");
    }
    for (k = 0; k < j; k++) {
      for (k1 = 0; k1 < q; k1++) {
        printf("%c", a[m]);
        m++;
      }
      printf("\n");
    }
  } else
    printf("-1\n");
  return 0;
}

```