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;
int n, i, j, k, a, b, poz;
int d[410][410], t[410][410], bc[410];
char s[410];
int main() {
  scanf("%d%d%d\n", &k, &a, &b);
  scanf("%s", s + 1);
  n = strlen(s + 1);
  d[0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < k; ++j) {
      if (d[i][j] == 0) continue;
      for (int k = a; k <= b; ++k) {
        d[i + k][j + 1] = 1;
        t[i + k][j + 1] = k;
      }
    }
  }
  if (d[n][k] == 0) {
    printf("No solution\n");
    return 0;
  }
  poz = n;
  for (int i = k; i; i--) {
    bc[i] = t[poz][i];
    poz -= t[poz][i];
  }
  poz = 1;
  for (int i = 1; i <= n; ++i) {
    bc[i] += bc[i - 1];
    printf("%c", s[i]);
    if (bc[poz] == i) {
      printf("\n");
      ++poz;
    }
  }
  return 0;
}
