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 $s$ consisting of $n$ 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 $s$ = "BBAABBBA" the answer is two-gram "BB", which contained in $s$ 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 $n$ ($2\le n\le 100$) — the length of string $s$. The second line of the input contains the string $s$ consisting of $n$ capital Latin letters.

Output

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $s$ 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;
}