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.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;

import static java.lang.System.out;

public class Main {

    Scanner scanner = new Scanner(System.in);

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


    void solve() {
        int k = scanner.nextInt(), a = scanner.nextInt(), b = scanner.nextInt();
        scanner.nextLine();

        String s = scanner.nextLine();

        int n = s.length();
        int p = n / k;
        if (n % k != 0) {
            p++;
        }

        if (a * k > n ||  b * k < n) {
            out.println("No solution");
        } else {
            if (n % k == 0) {
                for (int i = 0; i < k; i++) {
                    out.println(s.substring(i * p, i * p + p));
                }
            } else {
                int l = k;
                for (; l >= 1; l--) {
                    if ((n - l * (p - 1)) % p == 0) {
                        break;
                    }
                }

                int i = 0;
                for (i = 0; i < l; i++) {
                    out.println(s.substring(i * (p - 1), i * (p - 1) + (p - 1)));
                }
                for (i = 0; i < k - l; i++) {
                    out.println(s.substring(l * (p - 1) + i * p,  l * (p - 1) + i * p + p ));
                }
            }
        }

    }

}