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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define rep(i,n) for(int i=0;i<(n);++i)

int main()
{
    int n,m; cin >> n >> m;
    vector<int>team(n);
    int num_team = 1;
    rep(i,m){
        int a,b; cin >> a >> b; --a; --b;
        if ((team[a]==0)&&(team[b]==0)){ //まだどちらも友達組に入ってない
            team[a]=team[b]=num_team;
            num_team++;
        } else {
            if (team[a]==0) team[a]=team[b];
            if (team[b]==0) team[b]=team[a];
        }
    }

//rep(i,n) printf("%d: チーム %d\n",i,team[i]);

    vector<int>num_member(num_team);
    rep(i,n) num_member[team[i]]++;

//rep(i,num_team) printf("%d: メンバー %d\n",i,num_member[i]);

    int ans=0;
    rep(i,num_team) ans=max(ans,num_member[i]);
    cout << ans << endl;
    return 0;
}