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>
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 500 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
const int seed = 131;
int cases = 0;
int main() {
  int n, p, q;
  scanf("%d%d%d", &n, &p, &q);
  std::string str;
  std::cin >> str;
  for (int i = 0; i * p <= n; i++) {
    int rem = n - i * p;
    if (rem % q == 0) {
      printf("%d\n", i + rem / q);
      for (int j = 1; j <= i; j++)
        std::cout << str.substr((j - 1) * p, p) << std::endl;
      for (int j = 1; j <= rem / q; j++)
        std::cout << str.substr(i * p + (j - 1) * q, q) << std::endl;
      return 0;
    }
  }
  puts("-1");
  return 0;
}
