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 k, a, b, n, dp[209];
string s;
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> k >> a >> b >> s;
  n = s.size();
  for (int i = 0; i < k && n >= 0; i++) {
    n -= a;
    dp[i] = a;
  }
  if (dp[k - 1] == -1 || n < 0) {
    cout << "No solution";
    return 0;
  }
  for (int i = 0; i < k && n >= 0; i++) {
    while (dp[i] < b && n) n--, dp[i]++;
  }
  int sum = 0;
  for (int i = 0; i < k; i++) {
    sum += dp[i];
  }
  if (sum != s.size()) {
    cout << "No solution";
    return 0;
  }
  sum = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < dp[i]; j++) {
      cout << s[sum + j];
    }
    cout << "\n";
    sum += dp[i];
  }
}
