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>
using namespace std;
int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
int f, n, p, q, s[105], l;
char a[105];
bool DFS(int x) {
  if (x == n) return 1;
  if (x > n) return 0;
  if (DFS(x + p)) {
    l++, s[x + p] = 1;
    return 1;
  }
  if (DFS(x + q)) {
    l++, s[x + q] = 1;
    return 1;
  }
  return 0;
}
int main() {
  n = read(), p = read(), q = read();
  scanf("%s", a);
  for (int i = 0; i <= 100; i++) {
    for (int j = 0; j <= 100; j++)
      if ((i != 0 || j != 0) && i * p + j * q <= n &&
          n % (i * p + j * q) == 0) {
        printf("%d\n", n / (i * p + j * q) * (i + j));
        int t = 0, t1 = i, t2 = j;
        while (t < n) {
          if (t1)
            t += p, t1--;
          else if (t2)
            t += q, t2--;
          else
            t1 = i, t2 = j;
          s[t] = 1;
        }
        for (int i = 0; i < n; i++) {
          printf("%c", a[i]);
          if (s[i + 1]) printf("\n");
        }
        return 0;
      }
  }
  printf("-1\n");
  return 0;
}
