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;
using ll = long long;
using ull = unsigned long long;
string s;
int a, b;
vector<int> res;
bool solve(int i, int k);
int dp[205][205];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  memset(dp, -1, sizeof(dp));
  int k;
  cin >> k >> a >> b >> s;
  if (!solve(0, k)) {
    cout << "No solution\n";
  } else {
    int cur = 0;
    reverse(res.begin(), res.end());
    for (int i = 0; i < res.size();) {
      while (cur <= res[i]) {
        cout << s[cur];
        cur++;
      }
      i++;
      cout << '\n';
    }
  }
  return 0;
}
bool solve(int i, int k) {
  if (i == s.length()) {
    return dp[i][k] = (k == 0);
  }
  if (k == 0) {
    return dp[i][k] = false;
  }
  if (i > s.length() - a) {
    return dp[i][k] = false;
  }
  if (dp[i][k] != -1) {
    return dp[i][k];
  }
  bool flag = false;
  for (int x = i + a - 1; x <= min((int)s.length() - 1, i + b - 1); x++) {
    if (solve(x + 1, k - 1)) {
      res.push_back(x);
      flag = true;
      break;
    }
  }
  return dp[i][k] = flag;
}
