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>
using namespace std;
const int maN = 2e5+5;
int parent[maN];
int sz[maN];
int find_set(int v)
{
    if(v == parent[v])
        return v;
    return parent[v] = find_set(parent[v]);
}

void union_sets(int a, int b)
{
    int u = find_set(a);
    int v = find_set(b);
    if(u!=v)
    {
        if(sz[u] < sz[v])
        {
            swap(u, v);
        }
        parent[u] = v;
            sz[v] += sz[u];
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    
    for(int i = 0; i <= n; ++i)
    {
        parent[i] = i;
        sz[i] = 1;
    }
    for(int i = 0; i < m; ++i)
    {
        int a, b;
        cin >> a >> b;
        --a, --b;
        union_sets(a, b);
    }
    
    // for(int i = 1; i <= n; ++i)
    // {
    //     int aa = find_set(i);
    //     cout << parent[aa] << " ";
    // }
    int max_set = 1;
    for(int i = 0; i < n; ++i)
    {
        // cout << sz[i] << " ";
        if(max_set < sz[i])
            max_set = sz[i]; 
    }
    
    cout << max_set;

	return 0;
}