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 main() {
  int k, a, b, dp[205] = {0};
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int sz = s.size();
  for (int i = 1; i <= k; i++) {
    dp[i] = a;
  }
  int sum = a * k;
  if (a * k > sz || b * k < sz) {
    cout << "No solution" << endl;
    return 0;
  }
  s = "#" + s;
  int flag = 0;
  if (sum == sz) {
    flag = 1;
  } else {
    while (1) {
      flag = 0;
      for (int i = 1; i <= k; i++) {
        dp[i] = dp[i] + 1;
        sum++;
        if (sum == sz) {
          flag = 1;
          break;
        }
      }
      if (flag == 1) {
        break;
      }
    }
  }
  for (int i = 2; i <= k; i++) {
    dp[i] = dp[i - 1] + dp[i];
  }
  for (int i = 1; i <= dp[1]; i++) {
    cout << s[i];
  }
  cout << endl;
  for (int i = 2; i <= k; i++) {
    for (int j = dp[i - 1] + 1; j <= dp[i]; j++) {
      cout << s[j];
    }
    cout << endl;
  }
  return 0;
}
