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;
vector<int> adj[200005];
int visited[200005];
int curr=0;
int counter=0;
void explore(int v)
    {counter++;
    visited[v]=1;
    for(auto x:adj[v]){if(visited[x]==0){explore(x);}}
    }
int DFS(int n)
    {for(int i=0;i<n;i++){if(visited[i]==0){counter=0;explore(i);curr=max(curr,counter);}}
    return curr;
    }
int main()
    {ios::sync_with_stdio(0);
    cin.tie(0);
    int n,m;
    cin>>n>>m;
    int u,v;
    memset(visited,0,sizeof(visited));
    for(int i=0;i<m;i++)
        {cin>>u>>v;u--;v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
        }
    cout<<DFS(n)<<"\n";
    }