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.util.Scanner;

public class Competencia3 {

	public static void main(String[] args) {
			Scanner scan = new Scanner(System.in);
			int n = scan.nextInt();
			int p = scan.nextInt(), k = scan.nextInt();
			String string = scan.next();
			int resn = n;
			int divp = 0;
			int divk = 0;
			
			if (n % p == 0) {
				divp = n / p;
			}
			else if (n % k == 0) {
				divk = n / k;
			}
			else {
				resn -= p;
				for (int i = 1; resn > 0; i++) {
					if (resn % k == 0) {
						divp = i;
						divk = resn / k;
						break;
					}
					else {
						resn -= p;
					}
				}
			}
			if (divp == 0 && divk == 0) {
				System.out.println("-1");
			}
			else {
				System.out.println(divp + divk);
				String subs = string;
				for (int i = 0; i < divp; i++) {
					System.out.println(subs.substring(0, p));
					subs = subs.substring(p);
				}
				for (int i = 0; i < divk; i++) {
					System.out.println( subs.substring(0, k));
					subs = subs.substring(k);
				}
			}
			scan.close();
	}
}