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;
string s;
int a, b, k, dp[205][205], len, found = 0;
vector<int> v;
int solve(int leftLines, int leftLetters) {
  if (leftLines == 0)
    return dp[leftLines][leftLetters] = ((leftLetters == 0) ? 0 : INT_MIN);
  if (leftLetters == 0) return dp[leftLines][leftLetters] = INT_MIN;
  if (dp[leftLines][leftLetters] != -1) return dp[leftLines][leftLetters];
  int ans = INT_MIN;
  for (int i = a; i <= min(b, leftLetters); i++) {
    ans = max(ans, solve(leftLines - 1, leftLetters - i));
  }
  return dp[leftLines][leftLetters] = ans;
}
int main() {
  cin >> k >> a >> b >> s;
  memset(dp, -1, sizeof dp);
  len = s.size();
  int zzz = solve(k, len);
  if (zzz == INT_MIN) return cout << "No solution", 0;
  int x = k, y = len;
  while (y) {
    for (int i = a; i <= b; i++) {
      if (y - i >= 0) {
        if (dp[x - 1][y - i] == 0) {
          v.push_back(i);
          x = x - 1;
          y = y - i;
          break;
        }
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < v.size(); i++) {
    for (int j = 0; j < v[i]; j++) {
      cout << s[cnt++];
    }
    cout << endl;
  }
  return 0;
}
