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 n, limit, a, b, i, j, k;
  cin >> limit >> a >> b;
  string s;
  getline(cin, s);
  getline(cin, s);
  n = s.size();
  int dp[n + 1][limit + 1][2];
  for (i = 0; i <= n; i++)
    for (j = 0; j <= limit; j++) dp[i][j][0] = 0;
  dp[0][0][0] = 1;
  for (i = 1; i <= limit; i++) {
    for (k = 1; k <= n; k++) {
      for (j = a; j <= b; j++) {
        if (dp[k][i][0] == 0 && k >= j && dp[k - j][i - 1][0] == 1) {
          dp[k][i][0] = 1;
          dp[k][i][1] = j;
        }
      }
    }
  }
  if (dp[n][limit][0] == 0)
    cout << "No solution" << endl;
  else {
    int ara[limit], si = 0, r;
    i = limit, k = n;
    while (i != 0) {
      ara[si++] = dp[k][i][1];
      k -= dp[k][i][1];
      i--;
    }
    ara[0]--;
    for (i = 1; i < limit; i++) ara[i] += ara[i - 1];
    r = 0;
    for (i = 0; i < limit; i++) {
      if (i > 0) r = ara[i - 1] + 1;
      for (j = r; j <= ara[i]; j++) cout << s[j];
      cout << endl;
    }
  }
  return 0;
}
