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>
using namespace std;
string pal;
int a, b, k, div1, mod;
int main() {
  scanf("%d%d%d", &k, &a, &b);
  cin >> pal;
  div1 = pal.size() / k;
  mod = pal.size() % k;
  if (b * k < pal.size() or a * k > pal.size()) {
    printf("No solution");
    return 0;
  }
  for (int i = 0, pos = 0; i < k; i++) {
    int n = pos + div1;
    if (mod > 0) n++, mod--;
    if (n > b and i == 0) {
      printf("No solution");
      return 0;
    }
    for (int j = pos; j < n; j++) {
      printf("%c", pal[j]);
    }
    pos = n;
    printf("\n");
  }
  return 0;
}
