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;

/**
 * Created by Gleb on 01.04.2016.
 */
public class P612_A {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int p = scan.nextInt();
        int q = scan.nextInt();
        String s = scan.next();

        if (solve(n,p,q,s)){
             return;
        } else {
            if (solve(n,q,p,s)){
                return;
            }
        }

        System.out.println(-1);

    }

    private static boolean solve(int n, int p, int q, String s){

        int summ = 0;

        for (int k = 1; p * k <= n - n % p; k++) {
            int a = 0;
            a = (n - p * k);
            if (a % q == 0) {
                summ = k + a / q;
                System.out.println(summ);
                for (int i = 0; i < a / q; i++) {
                    System.out.println(s.substring(0, q));
                    s = s.substring(q, s.length());
                }
                for (int i = 0; i < k; i++) {
                    System.out.println(s.substring(0, p));
                    s = s.substring(p, s.length());
                }
                return true;
            }
        }
        return false;
    }
}