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: Runtime Error)

#include <bits/stdc++.h> 
#include <algorithm>  
#define ll long long int
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define fr(i,a,b) for(long long int i=a; i<b; i++)
#define pb push_back
#define mp make_pair
#define MOD 1000000007
using namespace std;
vector<ll>g[1000001];
ll arr[1000][1000];
ll vis[1000005]={0};
/*queue<ll>q ;
void bfs(ll s)
{
       vis[s]=1;
       q.push(s);
       ll curr=q.front(); vis[curr]=1;
       while(!q.empty())
       {
              curr=q.front(); q.pop(); // vis[curr]=1;
              for(ll i=0;i<g[curr].size();i++)
              {
                     ll child=g[curr][i];
                     if(vis[child]==0) 
                     { 
                            q.push(child); 
                            vis[child]=1;
                     }
              }
       }
}*/
ll co=0;
ll dfs(ll curr)
{
       vis[curr]=1;
       for(ll i=0;i<g[curr].size();i++)
       {
              ll child=g[curr][i];
              if(vis[child]==0)
              {
                     co++;
                     dfs(child);
              }
       }
       return co;
}
signed main() 
{
    fast
    ll n,m;
    cin>>n>>m;
    while(m--)
    {
           ll u,v;
           cin>>u>>v;
           if(arr[u][v]==0){g[u].pb(v);g[v].pb(u);}
           arr[u][v]=1;
    }
    ll c=0;
    vector<ll>v;
    for(ll i=1;i<=n;i++)
    {
           if(vis[i]==0)  {v.pb(dfs(i)+1); co=0; c++;}
    }
   // for(ll i=0;i<v.size();i++) cout<<v[i]<<" ";
    if(c==1) cout<<n<<" ";
    else
    {
           sort(v.begin(),v.end());
          cout<<v[0]+(v[v.size()-1]-v[0]);
    }
}