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)

/**
 *    author:  Mohamad Milhem
**/

#include <bits/stdc++.h>
#include <stdio.h>

using namespace std;
typedef long long ll;

#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mp make_pair
#define pb push_back
#define lp(i,s,f) for(ll i = s; i < ll(f); i++)
#define inF freopen("input.in", "r", stdin);
#define outF freopen("output.in", "w", stdout);
#define MOD ll(1000000007)
#define debug(x) cout << '[' << #x << " is: " << x << "] " <<endl;
#define decimalpoint cout << std::fixed << setprecision(5)
vector<int> adj[200005];
int sz;
bool visited[200005];


void dfs(int v){
	visited[v] = true;
	sz++;

	for (int i : adj[v]){
		if (visited[i] == false){
			dfs(i);
		}
	}

}

void solve(){
	int n, m;
	cin >> n >> m;

	for (int i = 0 ; i < m ; i++){
		int v,u;
		cin >> v >> u;
		adj[v].push_back(u);
		adj[u].push_back(v);
	}

	int maxx = 0;

	for (int i = 1 ; i <= n ; i++){
		sz = 0;
		if (visited[i] == false	)dfs(i);
		maxx = max(maxx , sz);
	}

	cout << maxx << endl;


}



int main()
{
    FAST;
    int t;
    t = 1;

    while(t--){
        solve();
    }
    return 0;
}