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;
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define sii set <ll>
#define vi vector <ll>
#define popcount(x) __builtin_popcountll(x)
#define mii map <ll,ll>
#define vpi vector <pair <ll,ll> >
#define sz(c) (int)c.size()
#define fr first
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define sc second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define mem0(a) memset(a,0,sizeof(a))
#define rep(i,a,n) for(ll i=a ; i<n ; i++)
#define ld long double
#define rall(a) (a).rbegin(),(a).rend()
#define INF 1000000000
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
 
 
 
ll power(ll b,ll e,ll m)
{
  if(b==0) return 0;
  if(e==0) return 1;
  if(e&1) return b*power(b*b%m,e/2,m)%m;
  return power(b*b%m,e/2,m);
}
ll power( ll b, ll e)
{
  if(b==0) return 0;
  if(e==0) return 1;
  if(e&1) return b*power(b*b,e/2);
  return power(b*b,e/2);
}  
bool isPowerOfTwo(ll x)
{
    // x will check if x == 0 and !(x & (x - 1)) will check if x is a power of 2 or not
    return (x && !(x & (x - 1)));
}

vi adj[200001];
ll vis[200001];
ll ans;

ll dfs(ll node)
{
  vis[node]=1;
  ans++;
  for(auto child:adj[node])
  {
    if(!vis[child])
    {
      dfs(child);
    }
  }
  return ans;
}

void solve()
{
  ll n,m,a,b;
  cin>>n>>m;
  rep(i,0,m)
  {
    cin>>a>>b;
    adj[a].pb(b);
    adj[b].pb(a);
  }
  ll out=0;
  rep(i,0,n+1)
  {
    ans=0;
    if(!vis[i])
    {
      ll mx=dfs(i);
      out=max(out,mx);
    }
  }
  cout<<out;
}


int main()
{
  fastio
  int tt=1;
  // cin>>tt;
  while(tt--)
  {
    solve();
    cout<<"\n";
  }
  return 0;
}