Problem:
Score: 300 points

Problem Statement

We have N voting papers. The i-th vote (1 \leq i \leq N) has the string S_i written on it.

Print all strings that are written on the most number of votes, in lexicographical order.

Constraints

1 \leq N \leq 2 \times 10^5

S_i (1 \leq i \leq N) are strings consisting of lowercase English letters.

The length of S_i (1 \leq i \leq N) is between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

Print all strings in question in lexicographical order.

Sample Input 1

7
beat
vet
beet
bed
vet
bet
beet

Sample Output 1

beet
vet

beet and vet are written on two sheets each, while beat, bed, and bet are written on one vote each. Thus, we should print the strings beet and vet.

Sample Input 2

8
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo

Sample Output 2

buffalo

Sample Input 3

7
bass
bass
kick
kick
bass
kick
kick

Sample Output 3

kick

Sample Input 4

4
ushi
tapu
nichia
kun

Sample Output 4

kun
nichia
tapu
ushi

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

import math
import queue
from collections import defaultdict

def readInt():
	return int(input())
def readInts():
	return list(map(int, input().split()))
def readChar():
	return input()
def readChars():
	return input().split()
def factorization(n):
	res = []
	if n%2==0:
		res.append(2)
	for i in range(3,math.floor(n//2)+1,2):
		if n%i==0:
			c = 0
			for j in res:
				if i%j==0:
					c=1
			if c==0:
				res.append(i)
	return res
def fact2(n):
	p = factorization(n)
	res = []
	for i in p:
		c=0
		z=n
		while 1:
			if z%i==0:
				c+=1
				z/=i
			else:
				break
		res.append([i,c])
	return res
def fact(n):#階乗
	ans = 1
	m=n
	for _i in range(n-1):
		ans*=m
		m-=1
	return ans
def comb(n,r):#コンビネーション
	if n<r:
		return 0
	l = min(r,n-r)
	m=n
	u=1
	for _i in range(l):
		u*=m
		m-=1
	return u//fact(l)
def combmod(n,r,mod):
	return (fact(n)/fact(n-r)*pow(fact(r),mod-2,mod))%mod
def printQueue(q):
	r=qb
	ans=[0]*r.qsize()
	for i in range(r.qsize()-1,-1,-1):
		ans[i] = r.get()
	print(ans)
def dq():
	return queue.deque()
class UnionFind():
	def __init__(self, n):
		self.n = n
		self.parents = [-1]*n

	def find(self, x): # root
		if self.parents[x]<0:
			return x
		else:
			self.parents[x] = self.find(self.parents[x])
			return self.parents[x]

	def union(self,x,y):
		x = self.find(x)
		y = self.find(y)

		if x==y:
			return

		if self.parents[x]>self.parents[y]:
			x,y = y,x

		self.parents[x]+=self.parents[y]
		self.parents[y]=x

	def size(self,x):
		return -1*self.parents[self.find(x)]

	def same(self,x,y):
		return self.find(x)==self.find(y)

	def members(self,x): # much time
		root = self.find(x)
		return [i for i in range(self.n) if self.find(i)==root]

	def roots(self):
		return [i for i,x in enumerate(self.parents) if x<0]

	def group_count(self):
		return len(self.roots())

	def all_group_members(self):
		return {r: self.members(r) for r in self.roots()} # 1~n
def bitArr(n):#ビット全探索
	x = 1
	zero = "0"*n
	ans = []
	ans.append([0]*n)
	for i in range(2**n-1):
		ans.append(list(map(lambda x:int(x),list((zero+bin(x)[2:])[-1*n:]))))
		x+=1
	return ans;
def arrsSum(a1,a2):
	for i in range(len(a1)):
		a1[i]+=a2[i]
	return a1
def maxValue(a,b,v):
	v2 = v
	for i in range(v2,-1,-1):
		for j in range(v2//a+1): #j:aの個数
			k = i-a*j
			if k%b==0:
				return i
	return -1


n = readInt()
d = defaultdict(int)
for i in range(n):
	d[readChar()]+=1

d = sorted(d.items(), key=lambda x:x[1], reverse=True)
maxCount = d[0][1]



for i in range(len(d)):
	if d[i][1]==maxCount:
		print(d[i][0])
	else:
		break