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>
#define boost_io ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define int long long
#define num(a) a-'0'
#define charr(a) (a - 'A' + 1)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define maxs(a,b) if(b>a)a=b
#define mins(a,b) if(b<a)a=b
#define debug(x) cerr<<"["<<#x<<":"<<x<<"] "
#define debug2(a,b) debug(a);debug(b)
#define debug3(a,b,c) debug2(a,b);debug(c)
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/(__gcd(a,b))
#define print(arr) for(auto it = arr.begin();it < arr.end();it ++){cout << *it << " ";}cout << ln;
#define all(a) (a).begin(), (a).end()
#define vi vector<long long>
#define v vector
#define p pair
#define pb push_back
#define mk make_pair
#define f first
#define s second
#define ln "\n"
typedef long double ld;
typedef double d;
using namespace std;
int modF=1e9+7;
int INF=1e11;

void solve(){
    int n,m;
    cin>>n>>m;
    v<vi> gr(n);
    int x,y;
    rep(i,0,m){
        cin>>x>>y;
        gr[x-1].pb(y-1);
        gr[y-1].pb(x-1);
    }
    int ans=0;
    vi vis(n,0);
    rep(i,0,n){
        if(vis[i]==0){
            vis[i]=1;
            queue<int> q;
            q.push(i);
            int counter=1;
            while(!q.empty()){
                int x=q.front();
                q.pop();
                for(int e:gr[x]){
                    if(vis[e]==0){
                        vis[e]=1;
                        q.push(e);
                        counter++;
                    }
                }
            }
            maxs(ans,counter);
        }
    }
    cout<<ans<<ln;
}

signed main(){
	boost_io;
	solve();
	return 0;
}