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
#include <bits/stdc++.h>
int main() {
  char str[10000], c;
  int k, a, b, case1, case2, i, main_case, len, x = 0, y, z, rem1, rem2;
  scanf("%d %d %d %s", &k, &a, &b, str);
  len = strlen(str);
  rem1 = len % k;
  case1 = len / k + 1;
  case2 = len / k;
  if (a * k > len || b * k < len) {
    printf("No solution\n");
  } else {
    for (i = 0; i < k; i++) {
      if (i < rem1)
        main_case = case1;
      else
        main_case = case2;
      while (main_case > 0) {
        printf("%c", str[x++]);
        main_case--;
      }
      printf("\n");
    }
  }
  return 0;
}
