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 TextSplitting {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int p = in.nextInt();
		int q = in.nextInt();
		double q2 = (double) q;
		double p2 = (double) p;
		String text = in.next();
		int length = text.length();

		if (p > n && q > n) {
			System.out.println(-1);
			return;
		} else if (n % p == 0) {
			System.out.println(n/p);
			for (int i = 0; i < length; i+=p) {
				System.out.println(text.substring(i, i+p));
			}
			return;
		} else if (n % q == 0) {
			System.out.println(n/q);
			for (int i = 0; i < length; i+=q) {
				System.out.println(text.substring(i, i+q));
			}
			return;
		} else {
			for (int i = 0; i < 50; i++) {
				for (int k = 0; k < 50; k++) {
					double rem = (n-i*p)/(k*q2);
					if (rem == 1) {
						System.out.println(i + (n-i*p)/q);
						for (int j = 0; j < i*p ; j+=p) {
							System.out.println(text.substring(j, j+p));
						}
						for (int j = i*p; j < length; j+=q) {
							System.out.println(text.substring(j, j+q));
						}
						return;
					}
					double rem2 = (n-i*q)/(k*p2);
					if (rem2 == 1) {
						System.out.println(i + (n-i*q)/p);
						for (int j = 0; j < i*q ; j+=q) {
							System.out.println(text.substring(j, j+q));
						}
						for (int j = i*q; j < length; j+=p) {
							System.out.println(text.substring(j, j+p));
						}
						return;
					}
				}
			}
		}
		System.out.println(-1);
	}

}