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.io.PrintWriter;
import java.util.Locale;
import java.util.Scanner;

public class E implements Runnable {
    private Scanner in;
    private PrintWriter out;

    private void solve() {
        int k = in.nextInt();
        int a = in.nextInt();
        int b = in.nextInt();
        char[] s = in.next().toCharArray();
        if (s.length < k * a || s.length > k * b) {
            out.println("No solution");
            return;
        }
        int left = s.length;
        int pos = 0;
        while (left - a <= (k - 1) * b) {
            for (int i = pos; i < pos + a; ++i) {
                out.print(s[i]);
            }
            out.println();
            left -= a;
            pos += a;
            --k;
            if (k == 0) {
                return;
            }
        }
        int cur = left % b == 0 ? b : left % b;
        for (int i = pos; i < pos + cur; ++i) {
            out.print(s[i]);            
        }
        out.println();
        pos += cur;
        --k;
        for (int i = 0; i < k; ++i) {
            for (int j = pos; j < pos + b; ++j) {
                out.print(s[j]);
            }
            out.println();
            pos += b;
        }
    }

    @Override
    public void run() {
        Locale.setDefault(Locale.US);
        in = new Scanner(System.in);
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

    public static void main(String[] args) {
        new E().run();
    }
}
