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, n, pre[201][201];
bool f[201][201];
string s;
vector<int> res;
int main() {
  cin >> k >> a >> b;
  cin >> s;
  n = s.size();
  s = " " + s;
  for (int i = a; i <= b; i++) {
    if (i <= n) {
      f[1][i] = true;
      pre[1][i] = 0;
    }
  }
  for (int i = 2; i <= k; i++) {
    for (int j = a; j <= b; j++) {
      for (int l = 1; l <= 200; l++) {
        if (i != k) {
          if (l + j <= n && f[i - 1][l] == true) {
            f[i][l + j] = true;
            pre[i][l + j] = l;
          }
        } else {
          if (l + j == n && f[i - 1][l] == true) {
            f[i][l + j] = true;
            pre[i][l + j] = l;
          }
        }
      }
    }
  }
  if (f[k][n] == true) {
    int x = k, y = n;
    while (x > 0) {
      res.push_back(y - pre[x][y]);
      y = pre[x][y];
      x--;
    }
    int id = 1;
    while (res.size()) {
      int u = res.back();
      while (u--) {
        cout << s[id];
        id++;
      }
      cout << "\n";
      res.pop_back();
    }
  } else {
    cout << "No solution";
  }
}
