Problem:
Score : 400 points

Problem Statement

There are N persons called Person 1 through Person N.

You are given M facts that "Person A_i and Person B_i are friends." The same fact may be given multiple times.

If X and Y are friends, and Y and Z are friends, then X and Z are also friends. There is no friendship that cannot be derived from the M given facts.

Takahashi the evil wants to divide the N persons into some number of groups so that every person has no friend in his/her group.

At least how many groups does he need to make?

Constraints

2 \leq N \leq 2\times 10^5

0 \leq M \leq 2\times 10^5

1\leq A_i,B_i\leq N

A_i \neq B_i

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
\vdots
A_M B_M

Output

Print the answer.

Sample Input 1

5 3
1 2
3 4
5 1

Sample Output 1

3

Dividing them into three groups such as \{1,3\}, \{2,4\}, and \{5\} achieves the goal.

Sample Input 2

4 10
1 2
2 1
1 2
2 1
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output 2

4

Sample Input 3

10 4
3 1
4 1
5 9
2 6

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,m,z,x,y,fa[N];
inline int find(int x)
{
    while(x!=fa[x]) x=fa[x]=fa[fa[x]];
    return x;
}
int num[N];
int ans;
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;++i) 
    	fa[i]=i;
    while(m--)
    {
        cin>>x>>y;
        int a=find(x),b=find(y);
        	fa[a]=b;
    }
    for (int i=1;i<=n;i++) num[find(i)]++;
    for (int i=1;i<=n;i++) ans=max(ans,num[i]);
    cout<<ans<<endl;
}