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

   void run() throws IOException
   {
      Scanner in = new Scanner(System.in);
      
      int n = in.nextInt();
      int a = in.nextInt();
      int b = in.nextInt();
      
      in.nextLine();
      
      String s = in.nextLine();
     
      int l = s.length();
      
      int[] A = new int[n];
      
      for(int i =0; i<n-1; i++){
          A[i] = a;
      }
      A[n-1] = l-(n-1)*a;
      
      if(A[n-1]<a){
          System.out.print("No solution");
          return;
      }
      
      for(int i= n-1; (A[i]> b)&(i>0); i--){
          int k = A[i] - b;
          A[i] = b;
          A[i-1] += k;
      }
      
     /* for(int i=0; i<n; i++){
          System.out.println(A[i]+" ");
      }*/
      
      
      if((A[0] >=a)&&(A[0]<=b)){
          int start = 0;
          for(int i=0; i<n; i++){
              System.out.println(s.substring(start, start+A[i]));
              start += A[i];
          }
      }else{
          System.out.print("No solution");
      }
      
   }
}