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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, n, m, z, l, r, curr = 0, x1 = 0, x2 = 0, city = 0, t,
                                        z1, z2, y1, y2, s, z3, w, s1, ans1, x,
                                        y, ans;
  cin >> k >> x >> y;
  vector<long long int> a(k, 0);
  string ch;
  cin >> ch;
  n = ch.length();
  if ((k * y) < n) {
    cout << "No solution\n";
    return 0;
  } else if ((k * x) > n) {
    cout << "No solution\n";
    return 0;
  }
  for (i = 0; i < k; i++) {
    a[i] = x;
    n = n - x;
  }
  for (i = 0; i < k; i++) {
    z = min(n, y - a[i]);
    n = n - z;
    a[i] = a[i] + z;
  }
  z = 0;
  n = ch.length();
  for (i = 0; i < k; i++) {
    for (j = 0; j < a[i]; j++) {
      cout << ch[z];
      z++;
    }
    cout << endl;
  }
}
