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 prime[100000];
vector<int> pr;
void calculateprime() {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= 100000; i++) {
    if (prime[i] == 0) {
      pr.push_back(i);
      for (long long j = i * 2; j <= 100000; j += i) {
        prime[j] = 1;
      }
    }
  }
}
void customrun() {}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool isprime(long long n) {
  for (long long i = 2; (long long)i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long fib(int n) {
  long long a = 0, b = 1, c, i;
  if (n == 0) return a;
  for (i = 2; i <= n; i++) {
    c = (a + b);
    a = b;
    b = c;
  }
  return b;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extendedgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extendedgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long findgrouplcm(vector<int> &arr, int n) {
  long long ans = arr[0];
  for (long long i = 1; i < n; i++) ans = (arr[i] * (ans / (gcd(arr[i], ans))));
  return ans;
}
long long powerwithmod(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool comp(const pair<long, long> &a, const pair<long, long> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
void solve();
int main() {
  clock_t begin = clock();
  fast();
  customrun();
  long long t = 1;
  for (long long i = (1); i < (t + 1); i += (1)) {
    solve();
  }
  clock_t end = clock();
  double time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
  return 0;
}
int res(vector<int> &v, map<int, int> &m, int l, int r) {
  if (r <= l) {
    return 0;
  }
  bool ok =
      (*upper_bound(v.begin() + l, v.begin() + r, 2 * v[l]) - 2 * v[l] <= 0);
  if (ok == true) return 0;
  int first = m[v[l]] + res(v, m, l + 1, r);
  int second = m[v[r]] + res(v, m, l, r - 1);
  return min(first, second);
}
void solve() {
  int k, a, b;
  cin >> (k) >> a >> b;
  string str;
  cin >> (str);
  int n = str.length();
  if (b * k < n or a * k > n) {
    cout << ("No solution");
    return;
  }
  int i = 0, len = n;
  while (k) {
    int l = n / k;
    if (l > b or l < a) {
      cout << ("No solution");
      return;
    }
    n -= l;
    for (int j = i; j < min(len, i + l); j++) {
      cout << str[j];
    }
    cout << "\n";
    i = i + l;
    --k;
  }
}
