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;
char s[250];
int main() {
  int k, a, b, m, i, j, sp, len;
  while (scanf("%d%d%d", &k, &a, &b) != EOF) {
    scanf("%s", s);
    m = strlen(s);
    len = m;
    if (m < a * k || m > b * k) {
      printf("No solution\n");
      continue;
    }
    sp = 0;
    for (i = 1; i <= k; i++) {
      m = len - sp;
      if (m > (k - i) * a + b) {
        for (j = sp; j < sp + b; j++) printf("%c", s[j]);
        printf("\n");
        sp += b;
        continue;
      }
      if (m > (k - i + 1) * a) {
        for (j = sp; j < len - (k - i) * a; j++) printf("%c", s[j]);
        printf("\n");
        sp = len - (k - i) * a;
        continue;
      }
      for (j = sp; j < sp + a; j++) printf("%c", s[j]);
      printf("\n");
      sp += a;
    }
  }
}
