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;
string str;
int k, a, b, dp[210][210];
int rec(int p, int k) {
  if (p == 0 && k == 0) return 1;
  if (p <= 0 || k < 0) return 0;
  int &temp = dp[p][k];
  if (temp != -1) return temp;
  for (int i = a; i <= b; i++) temp = max(temp, rec(p - i, k - 1));
  return temp;
}
void prin(int p, int k) {
  if (p == 0 && k == 0) return;
  if (p <= 0 || k < 0) return;
  for (int i = a; i <= b; i++) {
    if (rec(p - i, k - 1)) {
      prin(p - i, k - 1);
      for (int j = p - i; j <= p - 1; j++) cout << str[j];
      cout << endl;
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> k >> a >> b;
  cin >> str;
  memset(dp, -1, sizeof(dp));
  if (rec(str.size(), k))
    prin(str.size(), k);
  else
    cout << "No solution" << endl;
  return 0;
}
