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 A612 {
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		String ints[] = scan.nextLine().split(" ");
		String s = scan.nextLine();
		int n = Integer.parseInt(ints[0]);
		int p = Integer.parseInt(ints[1]);
		int q = Integer.parseInt(ints[2]);
		int i =0, j = 0;
		boolean f = false;
		for(; p*i <=n; i++)
		{
			if((n-p*i) % q == 0)
			{
				f= true;
				j = (n-p*i)/q;
				break;
			}
		}
		
		if(!f)
		{
			System.out.println("-1");
		}
		else
		{
			System.out.println(i+j);
			if(i != 0)
			{
				for(int k = 0; k<i; k++)
				{
					System.out.println(s.substring(p*k, p*k+p));
				}
			}
			if(j != 0)
			{
				for(int k=0; k<j; k++)
				{
					System.out.println(s.substring(q*k+p*i, q*k+p*i+q));
				}
			}
		}
		
	}
	
	public static int gcd(int a, int b)
	{
		return b ==0? a : gcd(b, a%b);
	}
}
