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, a, b;
string m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> a >> b;
  cin >> m;
  int leng = m.length();
  int j = leng / n;
  if (j >= a && j <= b) {
    if ((leng - j * n + j) > b) {
      int t = leng - j * n;
      if (j + 1 > b) {
        cout << "No solution" << '\n';
      } else {
        int q = (j + 1);
        int i = 0;
        while (i < q * t) {
          cout << m.substr(i, q) << '\n';
          i += q;
        }
        while (i < leng) {
          cout << m.substr(i, j) << '\n';
          i += j;
        }
      }
    } else {
      for (int i = 0; i < leng; i += j) {
        if (i == j * (n - 1)) {
          cout << m.substr(i, j + leng % n) << '\n';
          break;
        } else {
          cout << m.substr(i, j) << '\n';
        }
      }
    }
  } else {
    cout << "No solution" << '\n';
  }
  return 0;
}
