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;
long long bigmod(long long base, long long power, long long mod) {
  if (power == 0)
    return 1;
  else if (power % 2 == 1) {
    long long p1 = base % mod;
    long long p2 = (bigmod(base, power - 1, mod)) % mod;
    return (p1 * p2) % mod;
  } else if (power % 2 == 0) {
    long long p1 = (bigmod(base, power / 2, mod)) % mod;
    return (p1 * p1) % mod;
  }
}
long long dp[205][205];
long long k, a, b;
long long par[1233];
long long meet = -1;
long long call(long long rem, long long step, long long prev) {
  if (rem == 0 && step == k) {
    meet = prev;
    return 1;
  }
  if (step >= k || rem < 0) {
    return 0;
  }
  if (dp[rem][step] != -1) {
    return dp[rem][step];
  }
  long long ans = 0;
  for (int i = a; i <= b; i++) {
    if (rem - i >= 0) {
      long long x = call(rem - i, step + 1, i);
      if (x == 1) {
        par[rem - i] = rem;
      }
      ans |= x;
    }
  }
  return dp[rem][step] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long t = 1;
  while (t--) {
    memset(dp, -1, sizeof(dp));
    cin >> k >> a >> b;
    string s;
    cin >> s;
    long long n = s.size();
    call(n, 0, -1);
    if (meet == -1) {
      cout << "No solution" << endl;
      exit(0);
    }
    vector<long long> ans;
    ans.push_back(meet);
    while (1) {
      if (meet == n) break;
      ans.push_back(par[meet] - meet);
      meet = par[meet];
    }
    long long j = 0;
    for (int i = 0; i < ans.size(); i++) {
      for (int f = 0; f < ans[i]; f++) {
        cout << s[j++];
      }
      cout << endl;
    }
    exit(0);
  }
}
