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)


// Problem : D - Friends
// Contest : AtCoder - AtCoder Beginner Contest 177
// URL : https://atcoder.jp/contests/abc177/tasks/abc177_d
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define sort(a) sort(all(a))
#define ff first
#define ss second
#define vi vector<int>
#define pii pair<int, int>
#define vll vector<ll>
#define pll pair<ll,ll>
#define mod (int)1e9+7
#define FAST std::ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define repf(i,a,b) for(i=a;i<b;i++)
#define repb(i,a,b) for(i=a;i>=b;i--)
#define N 200005

void dfs(int x, int &cnt, vi g[], vector<bool> &vis)
{
	vis[x]=true;
	cnt++;
	int neigh=g[x].size();
	for(int i=0;i<neigh;i++)
	{
		if(!vis[g[x][i]])
		dfs(g[x][i],cnt,g,vis);
	}
}

int main()
{
	FAST
    int t=1;
    // cin>>t;
    while(t--)
    {
		int n,m;
		cin>>n>>m;
		
		vector<int> g[n+1];
		vector<bool>vis(n+1);		
		while(m--)
		{
			int x,y;
			cin>>x>>y;
			g[x].pb(y);g[y].pb(x);
		}
		
		int ma=-1;
		for(int u=1;u<=n;u++)
		{
			if(!vis[u])
			{
				int count=0;
				dfs(u,count,g,vis);
				ma=max(count,ma);
			}
		}
		cout<<ma;
    }
    return 0;
}