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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define pb(x)           push_back(x)
#define mkp(x,y)        make_pair(x,y)
#define L               length()
#define int             long long
#define bs              binary_search
#define mod             1e9+7
#define IOS             ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define all(v)          v.begin(),v.end()
#define allr(v)         v.rbegin(),v.rend()
#define mem0(a)         memset(a,0,sizeof(a))
#define mem1(a)         memset(a,-1,sizeof(a))
#define F               first
#define S               second
#define pii             pair<int,int>
#define vi              vector<int>
#define vs              size()
#define gcd(a,b)        __ gcd(a,b)
#define endl            '\n'
#define atoi            stoi
#define elasped_time    1.0 * clock() / CLOCKS_PER_SEC
#define si              set <int>
#define vpii            vector < pair <int,int> >
#define memf(a)         memset(a,false,sizeof(a))
#define memt(a)         memset(a,true,sizeof(a))
#define xxx             998244353
#define pi              3.141592653589
#define ninf            INT_MIN
#define inf             INT_MAX
#define sz(v)           ((int)(v).size())
#define rep(i,a,b)      for(int i=a;i<=b;i++)
int max(int a,int b){if(a>b){return a;}else{return b;}}
int min(int a,int b){if(a<b){return a;}else{return b;}}
int power(int b,int e)
{
    if(e==0)
    return 1;
    if(e%2)
    return ((b*power((b)*(b),(e-1)/2)));
    else
    return power((b)*(b),e/2);
}
int modpower(int b,int e,int q)
{
    int MOD=q;
    if(e==0)
    return 1;
    if(e%2)
    return ((b%MOD)*modpower((b%MOD)*(b%MOD),(e-1)/2,q))%MOD;
    else
    return modpower((b%MOD)*(b%MOD),e/2,q)%MOD;
}
void dpv(vi v)
{
    for(int i=0;i<v.vs;i++)
    {
        cout<<v[i]<<" ";
    }
    cout<<endl;
}
void dpv(vpii v)
{
    for(int i=0;i<v.vs;i++)
    {
        cout<<v[i].F<<" "<<v[i].S<<endl;
    }
}
void dpv(set <int> v)
{
    for(auto i:v)
    {
        cout<<i<<" ";
    }
    cout<<endl;
}
//////////////////////**TREE MOVES STARTS**//////////////////////////////
///// **TREE MOVES ENDS**////////////////////////////

////*SOLUTION TO THE QUESTION STARTS HERE*///////
vector < vi > adj(200006);
bool vis[200006];
int cnt=0;
void dfs(int x)
{
    cnt++;
    vis[x]=true;
    for(auto i:adj[x])
    {
        if(!vis[i])
        {
            dfs(i);
        }
    }
}
void oblivious()
{
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int x,y;
        cin>>x>>y;
        adj[x].pb(y);
        adj[y].pb(x);
    }
    int ma=-1;
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            cnt=0;
            dfs(i);
            ma=max(ma,cnt);
        }
    }
    cout<<ma<<endl;
}
signed main()
{
    IOS;
    /*#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif*/
    int t=1;
    //cin>>t;
    //int z=t;
    while(t--)
    {
        //cout<<"Case #"<<(z-t)<<": ";
        oblivious();
    }
    return 0;
}
//Editorial
/*

    */