#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
char str[maxn];
int main() {
  int n, q, p;
  while (~scanf("%d%d%d", &n, &p, &q)) {
    scanf("%s", str);
    int pos = -1;
    for (int i = 0; p * i <= n; i++) {
      if (n - p * i == 0 || (n - p * i) % q == 0) {
        pos = i;
        break;
      }
    }
    if (pos == -1) {
      puts("-1");
      continue;
    }
    int ans = pos + (n - pos * p) / q;
    printf("%d\n", ans);
    for (int i = 0; i < pos; i++) {
      for (int j = 0; j < p; j++) {
        printf("%c", str[i * p + j]);
      }
      puts("");
    }
    bool flag = false;
    for (int i = 0; i < (n - pos * p) / q; i++) {
      if (flag) {
        puts("");
      }
      flag = true;
      for (int j = 0; j < q; j++) {
        printf("%c", str[p * pos + i * q + j]);
      }
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>
using namespace std;
const int maxn = 110;
char str[maxn];
int main() {
  int n, q, p;
  while (~scanf("%d%d%d", &n, &p, &q)) {
    scanf("%s", str);
    int pos = -1;
    for (int i = 0; p * i <= n; i++) {
      if (n - p * i == 0 || (n - p * i) % q == 0) {
        pos = i;
        break;
      }
    }
    if (pos == -1) {
      puts("-1");
      continue;
    }
    int ans = pos + (n - pos * p) / q;
    printf("%d\n", ans);
    for (int i = 0; i < pos; i++) {
      for (int j = 0; j < p; j++) {
        printf("%c", str[i * p + j]);
      }
      puts("");
    }
    bool flag = false;
    for (int i = 0; i < (n - pos * p) / q; i++) {
      if (flag) {
        puts("");
      }
      flag = true;
      for (int j = 0; j < q; j++) {
        printf("%c", str[p * pos + i * q + j]);
      }
    }
  }
  return 0;
}

```