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
//package codeforcesSchool;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class E {

	static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);
	
	static String nextToken() throws IOException{
		while (st==null || !st.hasMoreTokens()){
			String s = bf.readLine();
			if (s == null)
				return null;
			st = new StringTokenizer(s);
		}
		
		return st.nextToken();
	}
	
	static int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	static String nextStr() throws IOException{
		return nextToken();
	}
	
	public static void main(String[] args) throws IOException{
		int k = nextInt(),
			a = nextInt(),
			b = nextInt(),
			c[] = new int[k];
		
		byte s[] = nextStr().getBytes();
		
		int l = s.length/k;
		if (l<a || l>b || l==b && s.length%k!=0){
			out.println("No solution");
			out.flush();
			return;
		}
		
		for (int i=0; i<k; i++)
			c[i] = l;
		int m = s.length%k;
		for (int i=0; i<k && m>0; i++){
			c[i]++;
			m--;
			if (c[i]<a || c[i]>b){
				out.println("No solution");
				out.flush();
				return;
			}
		}
		
		int p = 0;
		for (int i=0; i<k; i++){
			for (int j=0; j<c[i]; j++, p++)
				out.print((char)s[p]);
			out.println();
		}
		
		out.flush();
	}

}
