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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
constexpr int INF = 0x3f3f3f3f; constexpr ll LLINF = 0x3f3f3f3f3f3f3f3f;
#define db(x) { cerr << #x << " = " << x << endl; }
template <typename T> void _dbarr(T* a, size_t sz){ for(int i = 0; i < sz; i++) cerr << a[i] << " \n"[i == sz-1]; }
template <typename T> void _dbarr(vector<T> a, size_t sz){ for(int i = 0; i < sz; i++) cerr << a[i] << " \n"[i == sz-1]; }
#define dbarr(x, n) {cerr << #x << ": "; _dbarr((x),(n));}
#define all(x) begin(x), end(x)
#define pb push_back
#define eb emplace_back
#define mpr make_pair
#define fs first
#define sn second
const int MM = 2e5+5;
int n,m;
bool vis[MM];
vector<int> adj[MM];
int main()
{
	cin.tie(0)->sync_with_stdio(0);
    cin >> n >>m;
    for(int i = 0; i < m; i++)
    {
        int a,b;
        cin >> a >> b;
        adj[a].eb(b);
        adj[b].eb(a);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++)
    {
        if(!vis[i])
        {
            queue<int> q;
            q.push(i);
            vis[i] = 1;
            int cnt = 0;
            while(!q.empty())
            {
                cnt++;
                int u = q.front(); q.pop();
                for(int& v : adj[u])
                {
                    if(!vis[v])
                    {
                        vis[v] = 1;
                        q.push(v);
                    }
                }
            }
            ans = max(ans,cnt);
        }
    }
    cout << ans << "\n";
}