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 int M = 1e9 + 7;
const int N = 2e5;
long long po(int a, int b) {
  if (b == 0) return 1;
  long long ans = 1;
  if (b % 2 == 0) {
    ans = po(a, b / 2) % M;
    ans = (ans % M * ans % M) % M;
  } else {
    ans = po(a, (b - 1) / 2) % M;
    ans = (ans % M * ans % M * a) % M;
  }
  return ans % M;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int k, a, b;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int n = s.length();
  if (n / k >= a && n / k <= b) {
    int x = n / k;
    vector<string> v;
    k--;
    int h = k;
    int t = 0;
    while (k--) {
      string c;
      c.append(s, t, x);
      v.push_back(c);
      t += x;
    }
    string c;
    c.append(s, t, n - h * x);
    v.push_back(c);
    if (c.length() >= a && c.length() <= b) {
      for (auto i : v) {
        cout << i << '\n';
      }
    } else
      cout << "No solution\n";
  } else
    cout << "No solution\n";
  return 0;
}
