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 Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void err(istream_iterator<string> it) { cout << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ' ';
  err(++it, args...);
}
const int N = (int)2e5 + 5;
const int maxN = (int)1e6 + 6;
const long long Mod = (long long)1e9 + 7;
vector<int> box;
int len, k, a, b;
bool vis[201][201];
string s;
void call(int cur, int last) {
  if (cur > k) return;
  if (last == (int)s.size()) {
    if (cur != k) return;
    int cnt = 0;
    for (int i : box) {
      for (int j = cnt; j < i + cnt; ++j) cout << s[j];
      cout << '\n';
      cnt += i;
    }
    exit(0);
  }
  if (vis[cur][last]) return;
  vis[cur][last] = 1;
  for (int i = a; i <= b; ++i) {
    if (last + i <= (int)s.size()) {
      box.push_back(i);
      call(cur + 1, i + last);
      box.pop_back();
    }
  }
}
int main() {
  int test = 1, tc = 0;
  while (test--) {
    cin >> k >> a >> b;
    cin >> s;
    for (int i = a; i <= b; ++i) {
      if (i <= (int)s.size()) {
        box.push_back(i);
        call(1, i);
        box.pop_back();
      }
    }
    puts("No solution");
  }
  return 0;
}
