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
import java.io.*;

public class CF612A {
	
	public static void main(String[] args) throws IOException {
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);

		int n, p, q;
		String line[] = in.readLine().split(" ");
		n = Integer.parseInt(line[0]);
		p = Integer.parseInt(line[1]);
		q = Integer.parseInt(line[2]);
		
		String letters = in.readLine();

		boolean found = false;
		for (int i = 0; i <= n / p; i++) {
			if ((n - i*p) % q == 0) {
				int countp = i;
				int countq = (n-i*p) / q;
				out.println(countp+countq);
				for (int j = 0; j < i*p; j+=p) {
					out.println(letters.substring(j, j+p));
				}
				for (int j = i*p; j < n; j+=q) {
					out.println(letters.substring(j, j+q));
				}
				found = true;
				break;
			}
		}

		if (!found) out.println(-1);

		in.close();
		out.flush();

	}

}
