Codeforces 977B Two-gram（stl之string掉进坑）

Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.

You are given a string ss consisting of nn capital Latin letters. Your task is to find any two-gram contained in the given string as a substring(i.e. two consecutive characters of the string) maximal number of times. For example, for string ss = "BBAABBBA" the answer is two-gram "BB", which contained in ss three times. In other words, find any most frequent two-gram.

Note that occurrences of the two-gram can overlap with each other.

Input

The first line of the input contains integer number nn (2≤n≤1002≤n≤100) — the length of string ss. The second line of the input contains the string ss consisting of nn capital Latin letters.

Output

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string ss as a substring (i.e. two consecutive characters of the string) maximal number of times.

Examples

input

Copy

7
ABACABA

output

Copy

AB

input

Copy

5
ZZZAA

output

Copy

ZZ

Note

In the first example "BA" is also valid answer.

In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

思路：用map<string, int>维护一个最大值输出即可

坑点：string的加法为string = string + char* ， 不是string = char* + char* 或者char + char！

反正第一个加号前面必须得是string！

然后就是学一下map的一些操作了，注意map迭代器表示key和value的方式

代码：

#include<cstdio>
#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
typedef long long ll;
using namespace std;
const int maxn = 100000 + 100;

int main(){
    map<string, int> m;
    int n;
    scanf("%d", &n);
    getchar();
    string s;
    cin >> s ;
    //int len = s.length();
    string ans;
    int maxx = 0;
    for(int i = 0; i < n-1; i++){
    	string ss;
    	ss =ss+ s[i] + s[i+1];
    	//cout << ss<<endl;
    	if(!m.count(ss))m[ss] = 0;
    	m[ss]++;
    	//if(m[ss] > )
	}
	map<string, int>::iterator it;
	map<string, int>::iterator itt;

	for( it = m.begin(); it != m.end(); it++){
		if(it->second > maxx){
			itt = it;
			maxx = it->second;
			//cout << it->second << endl;
		}
	}
	cout <<itt->first;
    return 0;
}