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.BufferedReader;
import java.io.InputStreamReader;


public class m5 {
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] s = in.readLine().split(" ");
        int k = Integer.parseInt(s[0]);
        int a = Integer.parseInt(s[1]);
        int b = Integer.parseInt(s[2]);
        String inp = in.readLine();
        int n = inp.length();
        int max = b * k;
        int min = a * k;
        int linelen = n/k;
        int rem = n%k;
        int st = 0;
        if(!(n>=min && n<=max) || (linelen==b && rem>0) || (linelen>b)) {
            System.out.println("No solution");
            return;
        }
        for (int i = 0; i < k; i++) {
            int curr = linelen;
            if(curr<b && rem>0) {
                curr++;
                rem--;
            }
            System.out.println(inp.substring(st,st+curr));
            st = st+curr;
        }
   }
}



