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.*;
import java.math.BigInteger;
public class Main {
    
    public static void main(String[] args)throws IOException {
        Main principal = new Main();
        principal.input();
    }
    int [] array;
    int[][] dp;
    int a, b;
    private void input()throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int k, ini;
        String cad;
        boolean encontre = false;
        k = Integer.parseInt(st.nextToken());
        a = Integer.parseInt(st.nextToken());
        b = Integer.parseInt(st.nextToken());
        cad = br.readLine();
        dp = new int[k][cad.length()+1];
        int tam = cad.length();
        for (int i = a; i <= b; i++) {
            if (solve(k-1, tam-i, i)) {
                encontre = true;
                break;
            }
        }
        if(encontre){
            ini = 0;
            for (int i = 0; i < k; i++) {
                System.out.println(cad.substring(ini,dp[i][ini]+ini));
                ini = dp[i][ini]+ini;
            }
        } else
            System.out.println("No solution");
    }

    private boolean solve(int k, int i, int current){
        boolean answer = false;
        if(i >= 0)
            if (dp[k][i] != 0) {
                answer = false;
            }else
                if(k == 0){
                    if (i == 0) {
                        answer = true;
                        dp[k][i] = current;
                    }
                }else {
                    dp[k][i] = current;
                    for (int j = a; j <= b; j++) {
                        answer = solve(k-1, i-j, j);
                        if (answer) {
                             break;
                        } 
                    }
                }
        return answer;
    }

}
// a+b > c, a+c > b y b+c > a



