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 dp[][] = new int[n + 1][k + 1];

        for (int i = a; i <= Math.min(b, n); i++) {
            dp[i][1] = 1;
        }

        for (int i = 2; i <= k; i++) {
            for (int j = a; j <= b; j++) {
                for (int l = 1; l <= n; l++) {
                    if (dp[l][i - 1] != 0) {
                        if (l + j <= n) {
                            dp[l + j][i] = 1;
                        }
                    }
                }
            }
        }

        if (dp[n][k] != 1) {
            out.println("No solution");
            return;
        }

        int t = n;
        int cur = 0;
        for (int i = k; i >= 1; i--) {
            for (int j = a; j <= b ; j++) {
                if (t - j < 0) {
                    continue;
                }
                if (dp[t - j][i - 1] != 0) {
                    t = t - j;
                    out.println(s.substring(cur, cur + j));
                    cur = cur + j;
                    break;
                }
            }
        }
        out.println(s.substring(cur, s.length()));
    }

}