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 s;
int dp[251][251], a[251], n, i, j, z, k, l, r, nom;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> k >> l >> r;
  cin >> s;
  n = s.size();
  for (i = l; i <= r; i++)
    if (i <= n) dp[i][1] = -1;
  for (i = 1; i <= n - 1; i++)
    for (j = 1; j <= k - 1; j++)
      if (dp[i][j] != 0) {
        for (z = l; z <= r; z++)
          if (i + z <= n) dp[i + z][j + 1] = i;
      }
  if (dp[n][k] == 0) {
    cout << "No solution" << '\n';
    return 0;
  }
  a[1] = nom = n;
  for (i = 2; i <= k; i++) a[i] = nom = dp[nom][k - i + 2];
  a[k + 1] = 0;
  reverse(a + 1, a + 1 + k + 1);
  for (i = 1; i <= k; i++) {
    for (j = a[i]; j < a[i + 1]; j++) cout << s[j];
    cout << '\n';
  }
  return 0;
}
