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>
#define ff first
#define ss second
#define ll long long
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
const ll mod = 1e9 + 7;

void solve() {
   int n, m;
   cin >> n >> m;
   vector <vector <int>> adj(n + 1);
   for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
   }
   vector <bool> vis(n + 1);
   int comp = 0, ans = 0;
   function <void (int)> dfs = [&](int v) {
      vis[v] = true;
      comp++;
      for (auto u : adj[v]) {
         if (!vis[u])
            dfs(u);
      }
   };
   for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
         comp = 0;
         dfs(i);
         ans = max(ans, comp);
      }
   }
   cout << ans << '\n';
}

int main() {
#ifndef ONLINE_JUDGE
   freopen("inp.txt", "r", stdin);      freopen("outp.txt", "w", stdout);
#endif
   ios::sync_with_stdio(false);   cin.tie(NULL);   cout.tie(NULL);

   solve();

   return 0;
}