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;
template <typename T, typename T2>
T2 fac(T n) {
  return n < 2 ? 1 : fac<T, T2>(n - 1) * n;
}
template <typename T, typename T2, typename T3>
T2 fac(T n, T3 mod) {
  return n < 2 ? 1 : (fac(n - 1, mod) * n) % mod;
}
template <typename T>
T nb_combinaisons(T k, T n) {
  if (n < k) return nb_combinaisons<T>(n, k);
  return fac<T, long long int>(n) /
         (fac<T, long long int>(n - k) * fac<T, long long int>(k));
}
template <typename T, typename T2>
T nb_permutations(vector<T2> v) {
  T total = fac<T, T>(v.size());
  map<T2, T> m;
  bool distinct = true;
  for (long long unsigned c = 0; c < v.size(); c++) {
    m[v[c]]++;
    if (m[v[c]] - 1) distinct = false;
  }
  for (typename map<T2, T>::iterator it = m.begin(); it != m.end(); it++)
    total /= fac<T, T>(it->second);
  return total;
}
template <typename T, typename T2, typename T3>
T nb_permutations(vector<T2> v, T3 mod) {
  T total = fac<T, T, T3>(v.size(), mod);
  map<T2, T> m;
  bool distinct = true;
  for (long long unsigned c = 0; c < v.size(); c++) {
    m[v[c]]++;
    if (m[v[c]] - 1) distinct = false;
  }
  for (typename map<T2, T>::iterator it = m.begin(); it != m.end(); it++)
    total /= fac<T, T>(it->second);
  return total % mod;
}
string itoa_2(long long int n) {
  string s;
  stringstream ss;
  ss << n;
  ss >> s;
  while (s.size() != 9) s = '0' + s;
  return s;
}
int pgcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return pgcd(b, a % b);
}
string conversion(int n, int base) {
  string res = "";
  while (n != 0) {
    res = ((char)('0' + n % base)) + res;
    n /= base;
  }
  return res;
}
typedef struct speel {
  int percent;
  int dmg;
  int indice;
};
speel make_speel(int a, int b, int c) {
  speel s;
  s.percent = a;
  s.dmg = b;
  s.indice = c;
  return s;
}
bool comp(speel s1, speel s2) {
  if (s1.percent == s2.percent) {
    return s1.dmg > s2.dmg;
  }
  return s1.percent > s2.percent;
}
int main() {
  int k, a, b;
  string s;
  cin >> k >> a >> b >> s;
  if (k * b < s.size() || k * a > s.size()) {
    cout << "No solution";
    return 0;
  }
  int z = s.size() % k;
  int prec = 0;
  for (int c = 0; c < k; c++) {
    int prec2;
    if (c < z)
      prec2 = s.size() / k + 1;
    else
      prec2 = s.size() / k;
    cout << s.substr(prec, prec2) << endl;
    prec += prec2;
  }
}
