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 long long N = 2e5 + 10;
vector<long long> g[N];
set<pair<long long, long long> > CheckAgain;
set<long long> vis;

long long dfs(long long node) {
    if(vis.count(node))
        return 0;
    vis.insert(node);
    long long ans = 1;
    for(long long i = 0; i < g[node].size(); i++) {
        ans += dfs(g[node][i]);
    }
    return ans;
}

int main() {
 	long long n, m;
 	cin >> n >> m;
 	for(long long i = 1; i <= m; i++) {
        long long u, v;
        cin >> u >> v;
        if(CheckAgain.count({u, v}) || CheckAgain.count({v, u}))
            continue;
        g[u].push_back(v);
        g[v].push_back(u);
 	}
 	long long ans = 0;
 	for(long long i = 1; i <= n; i++) {
        if(!vis.count(i))
            ans = max(ans, dfs(i));
 	}
 	cout << ans << endl;
}