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本文介绍了一种通过优化路径查找根节点的方法来解决大规模数据集上寻找最大朋友集合的问题。该算法利用并查集实现，确保即使面对千万级别的数据也能高效运行。

Mr Wang wants some boys to help him with a project. Because the project is rather complex,the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

Sample Output

4
2

Hint

A and B are friends(direct or indirect), B and C are friends(direct or indirect), 
then A and C are also friends(indirect).
In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

题意：这道题目的意思就是选出最大的集合, 也就是人最多的集合, 另外, 如果所有点都是孤立点, 也就是说所有人都互不认识, 那么答案显然就是1

题解：此题数据大，故查找根节点要优化路径。

#include<iostream>
#include<algorithm>
#define N 100005
using namespace std;
int maxn; 
int set[N];
int num[N];
void Init_set(){
	for(int i=1;i<=N;i++){
		set[i]=i;
		num[i]=1;
	}
}
int find(int x){
	int r=x;
	while(set[r]!=r)
		r=set[r];
	int i=x;
	while(i!=r){
		int j=set[i];
		set[i]=r;
		i=j;
	}
	return r;
}
int merge(int a,int b){
	a=find(a);
	b=find(b);
	if(a!=b){
		set[b]=a;
		num[a]+=num[b];
		maxn=max(maxn,num[a]);
	}
	return maxn;
} 
int main(){
	int n;
	int a;
	int b;
	while(cin>>n){
		maxn=1;
		Init_set();
		for(int i=0;i<n;i++){
			cin>>a>>b;
			maxn=merge(a,b);
		}
		cout<<maxn<<endl;
	}
	return 0;
}