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;
bool can[205][205];
int main() {
  int k, a, b;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int n = s.length();
  can[0][n] = 1;
  for (int i = (0); i <= (k - 1); ++i) {
    for (int j = (0); j <= (n); ++j)
      if (can[i][j]) {
        for (int l = (a); l <= (b); ++l)
          if (j - l >= 0) can[i + 1][j - l] = 1;
      }
  }
  if (!can[k][0]) {
    puts("No solution");
    return 0;
  }
  vector<string> res;
  int u = k, v = 0;
  while (u > 0) {
    for (int i = (a); i <= (b); ++i) {
      if (can[u - 1][v + i]) {
        u--;
        v += i;
        string ans;
        for (int k = (1); k <= (i); ++k) {
          ans = s.back() + ans;
          s.pop_back();
        }
        res.push_back(ans);
        break;
      }
    }
  }
  reverse(res.begin(), res.end());
  for (auto x : res) {
    cout << x << endl;
  }
}
