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.*;
public class E {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int k=sc.nextInt(),a=sc.nextInt(),b=sc.nextInt();
        String n=sc.next();
        int[] ar=new int[k];
        Arrays.fill(ar, a);
        int diff=n.length()-k*a;
        if(diff<0) {
            System.out.println("No solution");
            return;
        }
        while(diff>0) {
            if(diff>=k) {
                if(ar[0]==b) {
                    System.out.println("No solution"); return;
                }
                Arrays.fill(ar,ar[0]+1);
                diff-=k;
            }
            if(diff==0) break;
            if(diff<k) {
                if(ar[0]==b) {
                    System.out.println("No solution"); return;
                }
                for(int i=0;i<diff;i++)
                    ar[i]++;
                break;
            }
        }
        int curidx=0;
        for(int i=0;i<ar.length;i++) {
            System.out.println(n.substring(curidx,curidx+ar[i]));
            curidx+=ar[i];
        }
    }
}
