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;
bool dp[205][205];
int opt[205][205];
vector<int> ans;
void backtrack(int x, int y) {
  if (x == 0 && y == 0) return;
  ans.push_back(opt[x][y]);
  backtrack(x - 1, y - opt[x][y]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int k, a, b, n;
  string s;
  cin >> k >> a >> b >> s;
  n = (int)s.length();
  dp[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = i; j <= n; j++) {
      for (int l = a; l <= b; l++) {
        if (j - l >= 0) {
          if (dp[i - 1][j - l]) {
            dp[i][j] = true;
            opt[i][j] = l;
          }
        }
      }
    }
  }
  if (!dp[k][n])
    cout << "No solution\n";
  else {
    backtrack(k, n);
    reverse(ans.begin(), ans.end());
    int x = 0;
    for (auto isi : ans) {
      cout << s.substr(x, isi) << "\n";
      x += isi;
    }
  }
  return 0;
}
