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> //Andrei Alexandru a.k.a Sho10
#define ll long long
#define double long double
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define all(a) (a).begin(), (a).end()
#define sz size
#define f first
#define s second
#define pb push_back
#define er erase
#define in insert
#define mp make_pair
#define pi pair
#define rc(s) return cout<<s,0
#define endl '\n'
#define mod 1000000007
#define PI 3.14159265359
#define MAXN 100005
#define INF 1000000005
#define LINF 1000000000000000005
#define CODE_START  ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
ll n,m,cnt=0;
vector<ll>g[200005];
map<ll,ll>viz;
vector<ll>v;
void dfs(ll node){
cnt++;
viz[node]=1;
for(auto it : g[node]){
    if(viz[it]==0){
        dfs(it);
    }
}
}
int32_t main(){
CODE_START;
cin>>n>>m;
for(ll i=1;i<=m;i++)
{
    ll x,y;
    cin>>x>>y;
    g[x].pb(y);
    g[y].pb(x);
}
for(ll i=1;i<=n;i++)
{
    if(viz[i]==0){
            cnt=0;
        dfs(i);
    v.pb(cnt);
    }
}
sort(v.begin(),v.end());
cout<<v[v.size()-1]<<endl;
}