// Original C++ solution (slower):
#include<iostream>

#include <algorithm>

using namespace std;

#define FOR(xxx,begin,num) for(int xxx=(begin);xxx<((begin)+(num));xxx++)

int main(){

  int N,cnt[200010];string S[200010],sw;

  cin >> N;

  FOR(i,0,N) cin >> S[i];

  sort(S, S + N);

  int tmp = 0,max = 0;

  cnt[0] = 1;

  FOR(i,1,N-1){

    if(S[tmp]==S[i]){cnt[tmp]++;}

    else{tmp = i;cnt[tmp]++;}

  }

  FOR(i,0,N){

    if(max < cnt[i]){

      max = cnt[i];

    }

  }

  FOR(i,0,N){

    if(cnt[i]==max) cout << S[i] << endl;

  }

  return 0;

}

// Optimized solution to the same problem (faster):
bool DBG = false;

// #pragma GCC optimize("Ofast")

// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")

// #pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>

//#include <boost/multiprecision/cpp_dec_float.hpp>

//#include <boost/multiprecision/cpp_int.hpp>

using namespace std;

using ll = long long;

using ld = long double;

//using i128 = __int128_t;

//using bint = boost::multiprecision::cpp_int

//using d1024 = boost::multiprecision::number<mp::cpp_dec_float<1024>>;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)

#define IFOR(i,a,b) for(int i=((b)-1);i>=(a);--i)

#define RPT(i,a,b) for(int i=(a);i<((a)+(b));++i)

#define IRPT(i,a,b) for(int i=((a)+(b)-1);i>=(a);--i)

#define ALL(x) x.begin(),x.end()

#define RALL(x) x.rbegin(),x.rend()

#define fs first

#define sd second

#define couts(x) cout << (x) << (" ")

#define coutn(x) cout << (x) << ("\n")

#define ncouts(x) numout(x),outst[outst_N++] = ' '

#define ncoutn(x) numout(x),outst[outst_N++] = '\n'

#define scouts(x) strout(x),outst[outst_N++] = ' '

#define scoutn(x) strout(x),outst[outst_N++] = '\n'

#define dcouts(x) if(DBG) couts(x)

#define dcoutn(x) if(DBG) coutn(x)

#define endl "\n"

#define psb push_back

#define ppb pop_back

#define eb emplace_back

#define lb lower_bound

#define ub upper_bound

#define LBIT(x,a) (((x)>>(a))&1LL)

#define IBIT(x,a) (((x)>>(a))&1)

#define BCOUNT(x) (__builtin_popcount(x))

template<typename T> std::istream &operator>>(std::istream &is, std::vector<T> &vec){ for (auto &v : vec) is >> v; return is; }

template<typename T1, typename T2> std::istream &operator>>(std::istream &is, std::pair<T1,T2> &p){is >> p.first >> p.second; return is; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::deque<T> &vec){ os << "deque["; for (auto v : vec) os << v << ","; os << "]"; return os; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }

template<typename T> std::ostream &operator<<(std::ostream &os, const std::unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }

template<typename T1, typename T2> std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }

template<typename TK, typename TV> std::ostream &operator<<(std::ostream &os, const std::map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }

template<typename TK, typename TV> std::ostream &operator<<(std::ostream &os, const std::unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }



template<class T> using V = vector<T>;

template<class T> using V2 = V<V<T>>;

template<class T> using V3 = V<V2<T>>;

template<class T> using V4 = V<V3<T>>;





char outst[20'000'000]; int outst_N = 0;

char outst_tmp[200];

template<class NUM >

void numout(NUM n){

  if(n<0) { n*=-1; outst[outst_N++] = '-';}

  if(n==0){ outst[outst_N++] = '0'; return;}

  int cnt = 0;

  while(n>0){

    outst_tmp[cnt++] = '0' + (n % 10);

    n /= 10;

  }

  IFOR(i,0,cnt){

    outst[outst_N++] = outst_tmp[i];

  }

}

void strout(std::string s){

  for(auto x: s){

    outst[outst_N++] = x;

  }

}



constexpr ll LINF = 1LL << 60;

constexpr int IINF = 1 << 28;

constexpr ll mod =  1'000'000'007;







void solve(){

  int n; cin >> n;

  map< string, int > mp; 

  FOR(i,0,n){

    string s; cin >> s;

    mp[s]++;

  }

  int cnt = 0;

  for(auto &x:mp){

    if(cnt<x.sd){ cnt = x.sd;}

  }

  for(auto &x:mp){

    if(cnt==x.sd) coutn( x.fs );

  }

}













int main(void){

  //std::cout << std::fixed << std::setprecision(20);

  cin.tie(0);

  //ios::sync_with_stdio(false);

  solve();

  //printf("%s", outst);

  return 0;

}








// Fastest known solution for this problem: