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 dp[201][201];
int k, a, b, si;
string x;
int cal(int idx, int rem) {
  if (rem > si) return 0;
  if (idx == k) {
    if (rem != si)
      return 0;
    else {
      dp[idx][rem] = 1;
      return 1;
    }
  }
  int &ret = dp[idx][rem];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = a; i <= b; i++) ret = max(cal(idx + 1, rem + i), ret);
  return ret;
}
void build(int idx, int rem) {
  if (idx == k) return;
  for (int i = a; i <= b; i++) {
    if (dp[idx][rem] == dp[idx + 1][rem + i]) {
      cout << x.substr(rem, i) << endl;
      build(idx + 1, rem + i);
      return;
    }
  }
}
int main() {
  cin >> k >> a >> b;
  cin >> x;
  memset(dp, -1, sizeof dp);
  si = x.size();
  cal(0, 0);
  if (cal(0, 0) == 0) {
    cout << "No solution";
    return 0;
  }
  build(0, 0);
  return 0;
}
