1. Title Description
It's a very simple question , Given $n$ A large integer , Find the maximum number of occurrences .

2. The basic idea
This problem can be done with hash ,ELFHash Wait for hash to pass .

3. Code 

 /* 1800 */

#include <iostream>

#include <sstream>

#include <string>

#include <map>

#include <queue>

#include <set>

#include <stack>

#include <vector>

#include <deque>

#include <bitset>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <ctime>

#include <cstring>

#include <climits>

#include <cctype>

#include <cassert>

#include <functional>

#include <iterator>

#include <iomanip>

using namespace std;

//#pragma comment(linker,"/STACK:102400000,1024000")
#define sti set<int>

#define stpii set<pair<int, int> >

#define mpii map<int,int>

#define vi vector<int>

#define pii pair<int,int>

#define vpii vector<pair<int,int> >

#define rep(i, a, n) for (int i=a;i<n;++i)

#define per(i, a, n) for (int i=n-1;i>=a;--i)

#define clr clear

#define pb push_back

#define mp make_pair

#define fir first

#define sec second

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

#define SZ(x) ((int)(x).size())

#define lson l, mid, rt<<1

#define rson mid+1, r, rt<<1|1
const int maxl = ;

const int MOD = ;

int H[MOD], cnt[MOD];

char s[maxl];

int n, ans;
inline int ELFHash(char *s) {

unsigned int h = ;

unsigned int g;

int i = ;
while (s[i]) {

h = (h<<) + s[i];

g = h & 0xf0000000L;

if (g) {

h ^= (g >> );

h &= ~g;

}

++i;

}
return h & 0x7fffffff;

}
inline void Hash(char *s) {

int i = ;
while (s[i]=='') ++i;

int k = ELFHash(s+i);

int t = k % MOD;
while (H[t]!=- && H[t]!=k)

t = (t + ) % MOD;
if (H[t] == -) {

cnt[t] = ;

H[t] = k;

} else if (++cnt[t] > ans) {

ans = cnt[t];

}

}
int main() {

ios::sync_with_stdio(false);

#ifndef ONLINE_JUDGE

freopen("data.in", "r", stdin);

freopen("data.out", "w", stdout);

#endif
while (scanf("%d",&n)!=EOF) {

memset(H, -, sizeof(H));

ans = ;

rep(i, , n) {

scanf("%s", s);

Hash(s);

}

printf("%d\n", ans);

}
#ifndef ONLINE_JUDGE

printf("time = %d.\n", (int)clock());

#endif
return ;

}

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