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 k, a, b;
vector<int> ans;
int dp[201][201];
bool solve(int l, int kp, vector<int> v) {
  if (dp[l][kp] != -1) return dp[l][kp];
  if (kp == k && l == 0) {
    ans = v;
    return true;
  }
  if (l <= 0 && kp != k) return false;
  v.push_back(l);
  bool x = false;
  for (int i = a; i <= b; i++) {
    x |= solve(l - i, kp + 1, v);
  }
  dp[l][kp] = x;
  return x;
}
int main(int argc, char const *argv[]) {
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int len = s.length();
  memset(dp, -1, sizeof(int) * 201 * 201);
  vector<int> v;
  solve(len, 0, v);
  if (ans.size() == 0) {
    cout << "No solution";
    return 0;
  }
  ans.push_back(0);
  s = "x" + s;
  string str = "";
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size() - 1; i++) {
    for (int j = ans[i] + 1; j <= ans[i + 1]; j++) {
      str += s[j];
    }
    str += "\n";
  }
  cout << str;
  return 0;
}
