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.

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)

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

4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

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.
/*两个朋友合并和的时候，需要将他们每个人的朋友数也一起合并。 
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define M 10000005
int pre[M];
int sum[M];
int k;
int find(int N)
{
	return pre[N]==N?N:pre[N]=find(pre[N]);
}
void join(int x,int y)
{
	int fx=find(x),fy=find(y);
	if(fx==fy)
	return;
	if(fx!=fy)
	{
		if(sum[fx]<sum[fy])
		{
			pre[fx]=fy;
			sum[fy]+=sum[fx];
			k=max(k,sum[fy]);
		}
		else
		{
			pre[fy]=fx;
			sum[fx]+=sum[fy];
			k=max(k,sum[fx]);
		}
	}
}
int main()
{
	int i,n,a,b;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<M;i++)//sum[M]初始化均为1； 
		{
			pre[i]=i;
			sum[i]=1;
		}
		if(n==0)
		printf("1\n");
		else
		{
			k=0;
		    while(n--){
			scanf("%d %d",&a,&b);
			join(a,b);}	
		    printf("%d\n",k);
	    }
	}
	return 0;
}