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;
const int N = 222;
char dp[N][N][N];
int n, k, a, b;
string s;
bool thereIsAnAnser() {
  return ((n / k >= a && n / k < b) ||
          ((n / k >= a && n / k == b) && (n % k == 0)));
}
int main() {
  cin >> k >> a >> b >> s;
  n = s.size();
  if (thereIsAnAnser()) {
    int AnsIdx;
    for (int Test = 0; Test <= b - a; Test++) {
      for (int len = a; len <= b; len++) {
        int Row = k - 1, idx = n, LEN = len;
        while (idx >= 0) {
          for (int dpLtr = 0, strLtr = idx - LEN; strLtr < idx;
               dpLtr++, strLtr++)
            dp[Test][Row][dpLtr] = s[strLtr];
          idx -= LEN;
          Row--;
          if (Row == 0 && (idx - LEN < 0 || idx <= b)) LEN = idx;
          if (Row < 0 && idx > 0)
            break;
          else if (Row < 0 && idx == 0) {
            AnsIdx = Test;
            goto ans;
          }
        }
      }
    }
  ans:
    for (int Row = 0; Row < k; Row++) cout << dp[AnsIdx][Row] << endl;
  } else if (!thereIsAnAnser())
    puts("No solution");
  return 0;
}
