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;
void solve() {
  int k, a, b;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  int l = s.length();
  double c = (double)l / k;
  int c1 = floor(c);
  int c2 = ceil(c);
  int flag1 = 0, flag2 = 0;
  if ((l - (c1 * (k - 1))) > b || (l - (c1 * (k - 1))) < a) {
    flag1 = 1;
  }
  if ((l - (c2 * (k - 1))) > b || (l - (c2 * (k - 1))) < a) {
    flag2 = 1;
  }
  if ((flag1 == 1 || c1 < a || c1 > b) && (flag2 == 1 || c2 < a || c2 > b)) {
    cout << "No solution";
    return;
  }
  if (flag1 == 0) {
    int i1 = 0;
    int kk = 0;
    while (i1 <= l) {
      kk++;
      for (long long int i = i1; i < min(i1 + c1, l); i++) {
        cout << s[i];
      }
      i1 += c1;
      if (kk == k) {
        for (long long int i = i1; i < l; i++) {
          cout << s[i];
        }
        return;
      }
      printf("\n");
    }
  } else {
    int i1 = 0;
    int kk = 0;
    while (i1 <= l) {
      kk++;
      for (long long int i = i1; i < min(i1 + c2, l); i++) {
        cout << s[i];
      }
      i1 += c2;
      if (kk == k) {
        for (long long int i = i1; i < l; i++) {
          cout << s[i];
        }
      }
      printf("\n");
    }
  }
}
int main() { solve(); }
