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;

typedef long long ll;

#define pb push_back
#define mp make_pair
#define f first
#define s second
#define all(a) a.begin(), a.end()

const int maxn = 200010;
const int MOD = 1e9+7;
const int INF = 1e9;
const ll LINF = 1e18;
int n, m, a[maxn];
string s, ss;

vector<int> adj[maxn];
bool vis[maxn];
int nodes;

void dfs(int s)
{
    vis[s]=true;
    nodes++;
    for(auto u : adj[s])
    {
        if(vis[u])continue;
        dfs(u);
    }
}

void solve()
{
    cin >> n >> m;
    fill(vis, vis+n+2, false);
    for(int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    int ans=0;
    for(int i = 1; i <= n; i++)
    {
        if(vis[i]) continue;
        nodes=0;
        dfs(i);
        ans=max(ans, nodes);
    }
    cout << ans << "\n";
}

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t=1;
    //cin >> t;
    while(t--) solve();
}