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.*;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            int k = in.nextInt();
            int a = in.nextInt();
            int b = in.nextInt();
            String s = in.next();
            int per = s.length() / k;
            int[] len = new int[k];
            for (int i = 0; i < k; i++) {
                len[i] = per;
            }
            int left = s.length() - per * k;
            for (int i = 0; left > 0; i = (i + 1) % k) {
                len[i]++;
                left--;
            }
            if(ok(len, a, b)) {
                int index = 0;
                for(int i=0; i<k; i++) {
                    System.out.println(s.substring(index, index += len[i]));
                }
            } else {
                System.out.println("No solution");
            }
        }
        in.close();
    }

    private static boolean ok(int[] v, int a, int b) {
        for(int i=0; i<v.length; i++) {
            if(v[i] < a || v[i] > b) {
                return false;
            }
        }
        return true;
    }

}
 		     	   	 	 	  		 	  			 		