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, f = 0;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int n = s.length();
  if (a * k > n || b * k < n) {
    cout << "No solution" << endl;
    return 0;
  }
  int dp[k];
  for (int i = 0; i < k; ++i) dp[i] = a;
  int sum = k * a;
  if (sum == n) {
    f = 1;
  } else {
    while (1) {
      f = 0;
      for (int i = 0; i < k; ++i) {
        dp[i]++;
        sum++;
        if (sum == n) {
          f = 1;
          break;
        }
      }
      if (f) break;
    }
  }
  int x = 0;
  for (int i = 1; i < k; ++i) dp[i] += dp[i - 1];
  for (int i = 0; i < k; ++i) {
    for (int j = x; j < dp[i]; ++j) {
      cout << s[j];
    }
    cout << endl;
    x = dp[i];
  }
  return 0;
}
