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.*;
import java.util.*;

public class ObeziankaAnfisa {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer in;
	static PrintWriter out = new PrintWriter(System.out);
	
	public static String nextToken() throws IOException{
		while (in == null || !in.hasMoreTokens()){
			in = new StringTokenizer(br.readLine());
		}
		
		return in.nextToken();
	}
	
	public static int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	public static double nextDouble() throws IOException{
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String[] args) throws IOException{
		int k = nextInt();
		int a = nextInt();
		int b = nextInt();
		String S = nextToken();
		
		int n = S.length();
		
		if (k * a > n || k * b < n){
			out.println("No solution");
			out.close();
			return;
		}
		
		int len = n / k;		
		int[] ans = new int[k];
		Arrays.fill(ans, len);
		for (int i = 0; i < n % k; i++) {
			ans[i]++;
		}
		
		int s = 0;
		for (int i = 0; i < k; i++) {
			out.println(S.substring(s, s + ans[i]));
			s += ans[i];
		}
		
		out.close();
	}
}
