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 <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
int fa[200005];
int son[200005];
int ans = 0;
int N, M, A, B;
int Find(int x)
{
    if (fa[x] == x)
        return x;
    else
    {
       return fa[x] = Find(fa[x]);
    }
}
void cal(int x, int y)
{
    int xx = Find(x);
    int yy = Find(y);
    //cout << xx << " --" << yy << endl;
    if (xx == yy)
        return;
    else
    {
        fa[xx] = yy;
        son[yy] += son[xx];
    }
}
void init()
{
    for (int i = 0; i <= N; i++)
    {
        son[i] = 1;
        fa[i] = i;
    }
}
int main()
{

    cin >> N >> M;
    init();
    for (int i = 1; i <= M; i++)
    {
        cin >> A >> B;
        cal(A, B);
    }
     for (int i = 1; i <= N; i++)
     {
        ans = max(ans,son[i]);
     }
    cout << ans << endl;
    return 0;
}