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;
string a, b;
long long arr[200009];
int main() {
  long long i, j, k, n, m, ans = 0, p, q, pt = -1, qt;
  scanf("%lld %lld %lld", &n, &p, &q);
  for (i = 0; i <= 100; i++) {
    for (j = 0; j <= 100; j++) {
      if ((p * i + q * j) == n) {
        pt = i;
        qt = j;
        break;
      }
    }
  }
  char str[1000];
  scanf("%s", &str);
  if (pt == -1)
    printf("-1");
  else {
    printf("%lld\n", pt + qt);
    int ind = 0;
    for (i = 0; i < pt; i++) {
      for (j = 0; j < p; j++) {
        printf("%c", str[ind++]);
      }
      printf("\n");
    }
    for (i = 0; i < qt; i++) {
      for (j = 0; j < q; j++) {
        printf("%c", str[ind++]);
      }
      printf("\n");
    }
  }
  return 0;
}
