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;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  s = '_' + s;
  int n = s.size();
  bool dp[k][n];
  memset(dp, false, sizeof(dp));
  int pre[k][n];
  memset(pre, -1, sizeof(pre));
  for (int i = a; i <= n and i <= b; i++) {
    dp[0][i] = true;
  }
  for (int i = 1; i < k; i++) {
    for (int j = 1; j < n; j++) {
      if (j >= a) {
        for (int k = a; k <= b; k++) {
          if (j - k > 0) {
            if (dp[i - 1][j - k] == true) {
              pre[i][j] = k;
              dp[i][j] = true;
            }
          }
        }
      }
    }
  }
  if (!dp[k - 1][n - 1]) {
    cout << "No solution";
    return 0;
  }
  vector<int> sol;
  sol.push_back(n - 1);
  int i = k - 1;
  int j = n - 1;
  while (i > 0) {
    sol.push_back(j - pre[i][j]);
    j -= pre[i][j];
    i--;
  }
  reverse(sol.begin(), sol.end());
  int start = 1;
  for (auto end : sol) {
    for (int i = start; i <= end; i++) {
      cout << s[i];
    }
    cout << endl;
    start = end + 1;
  }
  return 0;
}
