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 n;
int p, q, temp, ans;
string s;
int main() {
  scanf("%d%d%d\n", &n, &p, &q);
  cin >> s;
  for (int i = 0; i <= n / p; i++)
    for (int j = 0; j <= n / q; j++)
      if (i * p + j * q == n) {
        printf("%d\n", i + j);
        if (i && j) {
          for (int k = 1; k <= i; k++) {
            for (int l = 0 + (k - 1) * p; l < k * p; l++) printf("%c", s[l]);
            puts("");
          }
          for (int k = 1; k <= j; k++) {
            for (int l = i * p + (k - 1) * q; l < i * p + k * q; l++)
              printf("%c", s[l]);
            puts("");
          }
        }
        if (!i && j) {
          for (int k = 1; k <= j; k++) {
            for (int l = 0 + (k - 1) * q; l < k * q; l++) printf("%c", s[l]);
            puts("");
          }
        }
        if (i && !j) {
          for (int k = 1; k <= i; k++) {
            for (int l = 0 + (k - 1) * p; l < k * p; l++) printf("%c", s[l]);
            puts("");
          }
        }
        return 0;
      }
  puts("-1");
}
