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

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

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.
 
 題意：王老師要找一些男生幫助他完成一項工程。要求最後挑選出的男生之間都是朋友關系，可以說直接的，也可以是間接地。問最多可以挑選出幾個  男生（最少挑一個)。
 代碼：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int num[1000000],par[1000000];
 6 void init()
 7 {
 8     for(int i=1;i<1000000;i++)
 9      { par[i]=i;
10        num[i]=1;    
11      }
12 }
13 int find(int x)
14 {
15     if(par[x]!=x)
16       return par[x]=find(par[x]);
17     return x;
18     
19 }
20 void unite(int a,int b)
21 {
22     int fa=find(a);
23     int fb=find(b);
24     if(fa!=fb)
25     {par[fa]=fb;
26     num[fb]+=num[fa];  //這個地方註意，第一次num【fb】寫成了b，
27         
28     }
29 }
30 int main()
31 {
32     int a,b,n; 
33     while(scanf("%d",&n)!=EOF)
34     {if(n==0) 
35       {printf("1\n");
36        continue;
37           
38       } 
39      init();
40      int max=0;
41      while(n--)
42       {
43        scanf("%d %d",&a,&b);
44        if(a>max) max=a;
45        if(b>max) max=b;    
46        unite(a,b);
47       }
48       int m=0;
49      for(int i=1;i<=max;i++)
50        {if(num[i]>m)
51           m=num[i];
52            
53        } 
54      printf("%d\n",m);  
55      
56      
57         
58     }
59     return 0;
60 }

hdu--1856 More is better