Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into k lines not shorter than a and not longer than b, for the text to resemble human speech more. Help Anfisa.

Input

The first line contains three integers k, a and b (1 ≤ k ≤ 200, 1 ≤ a ≤ b ≤ 200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.

Output

Print k lines, each of which contains no less than a and no more than b symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). 

Examples

Input

3 2 5
abrakadabra


Output

ab
rakad
abra


Input

4 1 2
abrakadabra


Output

No solution

import java.util.Scanner;
import java.util.StringTokenizer;

public class E44 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String line = in.nextLine();
        StringTokenizer st = new StringTokenizer(line);
        int k = Integer.parseInt(st.nextToken());
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        line = in.nextLine();
        int n = line.length();
        int len = n / k;
        int rem = n % k;
        String[] res = new String[k];
        for (int i = 0 ; i < k ; i++) {
            if (rem > 0) {
                res[i] = line.substring(0, len + 1);
                rem--;
                line = line.substring(len + 1);
            } else {
                res[i] = line.substring(0, len);
                line = line.substring(len);
            }
        }
        boolean ok = true;
        for (int i = 0 ; i < k ; i++)
            ok &= (res[i].length() >= a && res[i].length() <= b);
        if (!ok)
            System.out.println("No solution"); else {
            for (int i = 0 ; i < k ; i++)
                System.out.println(res[i]);
            }
    }
}
