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 long long MOD = (long long)1e9 + 7;
const int N = 205, inf = 1e9 + 5;
long long add(long long x, long long y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
long long mult(long long x, long long y) { return (x * y) % MOD; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int K, a, b;
  cin >> K >> a >> b;
  string s;
  cin >> s;
  bool dp[N][N];
  int chars_in_kth_line[N][N];
  for (int n = a; n <= b; ++n) {
    dp[1][n] = true;
    chars_in_kth_line[1][n] = n;
  }
  for (int k = 2; k <= K; ++k) {
    for (int n = 1; n <= s.size(); ++n) {
      for (int kth_chars = a; kth_chars <= n && kth_chars <= b; ++kth_chars) {
        if (dp[k - 1][n - kth_chars] > dp[k][n]) {
          dp[k][n] = dp[k - 1][n - kth_chars];
          chars_in_kth_line[k][n] = kth_chars;
        }
      }
    }
  }
  if (dp[K][s.size()]) {
    int curr_n = s.size();
    vector<string> ans;
    for (int k = K; k >= 1; --k) {
      ans.push_back(s.substr(curr_n - chars_in_kth_line[k][curr_n]));
      s.erase(curr_n - chars_in_kth_line[k][curr_n]);
      curr_n -= chars_in_kth_line[k][curr_n];
    }
    reverse(begin(ans), end(ans));
    for (auto x : ans) cout << x << endl;
  } else {
    cout << "No solution" << endl;
  }
}
