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 A
{
	public void solve()
	{
		Scanner cin = new Scanner(System.in);
		int N = cin.nextInt();
		int p = cin.nextInt();
		int q = cin.nextInt();
		String s = cin.next();
		boolean find = false;
		for(int pi = 0;pi <= s.length() / p;pi++)
		{
			for(int qi = 0;qi <= s.length() / q;qi++)
			{
				//System.out.println(pi + " " + qi);
				if(p * pi + q * qi == N)
				{
					System.out.println(pi + qi);
					for(int i = 0;i < pi;i++)
					{
						StringBuilder sb = new StringBuilder();
						for(int j = 0;j < p;j++)
						{
							sb.append(s.charAt(i * p + j));
						}
						System.out.println(sb.toString());
					}
					for(int i = 0;i < qi;i++)
					{
						StringBuilder sb = new StringBuilder();
						for(int j = 0;j < q;j++)
						{
							sb.append(s.charAt(pi * p + i * q + j));
						}
						System.out.println(sb.toString());
					}
					find = true;
					break;
				}
				if(find)break;
			}
			if(find)break;
		}
		if(!find)System.out.println(-1);

	}
	public static void main(String[] args)
	{
		new A().solve();
	}
}