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;
priority_queue<long long> cancel;
long long arr[200005];
long long brr[200005];
long long dp[3005][3005];
int main() {
  long long test = 1;
  for (long long z = 0; z < test; z++) {
    long long n, mini, maxi;
    cin >> n >> mini >> maxi;
    string str;
    cin >> str;
    if (str.length() < n * mini || str.length() > n * maxi) {
      cout << "No solution" << endl;
    } else {
      arr[0] = maxi;
      for (long long i = 1; i < n; i++) {
        arr[i] = maxi;
      }
      long long sum = n * maxi;
      long long i = n - 1;
      while (sum > str.length()) {
        long long diff = min((long long)(sum - str.length()), maxi - mini);
        arr[i] -= diff;
        sum -= diff;
        i--;
      }
      long long j = 0;
      for (long long i = 0; i < n; i++) {
        string temp = "";
        for (long long k = 0; k < arr[i]; k++) {
          temp += str[j];
          j++;
        }
        cout << temp << endl;
      }
    }
  }
}
