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;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define si(n) scanf("%d",&n);
#define PI acos(-1.0)
#define sl(n) scanf("%lld",&n);
#define sf(n) scanf("%lf",&n);
#define ss(n) scanf("%s",n);
#define for0(i,n) for(ll i=0;i<n;i++)
#define for1(i,n) for(ll i=1;i<=n;i++)
#define forc(i,j,n) for(ll i=j;i<n;i++)
#define pb push_back
#define sor(n) sort(n.begin(),n.end())
#define sorcmp(n) sort(n.begin(),n.end(),cmp)
#define rev(n) reverse(n.begin(),n.end())
#define pi(x)  printf("%d",x);
#define pl(x)  printf("%lld",x);
#define pf(x)  printf("%.10lf",x);
#define ps(x)  printf("%s",x);
#define en() printf("\n");
#define random_shuffle(r...)random_shuffle(r,[](int _){return rand()%_;})
#define SQR(a) (a)*(a)
#define sp(x) fixed<<setprecision(x)
#define mod 1000000007
#define endl '\n'
#define pii pair<int,int>
#define pll pair<ll,ll>
#define bug cout<<"Here"<<endl
void fastio(){ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);}
template <typename T> T POW(T B,T P){ if(P==0) return 1; if(P&1) return B*POW(B,P-1);  else return SQR(POW(B,P/2));}
ll powmod(ll a, ll b){ll ans = 1;while(b){if(b&1) ans = ans*a%mod;a = a*a%mod;b >>= 1;}return ans;}


#define mxn 200005

vector<int>g[mxn];
int vis[mxn];


int bfs(int x)
{
    vis[x]=1;
    int cnt=1;

    queue<int>q;
    q.push(x);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();

        for(auto v:g[u])
        {
            if(vis[v]) continue;
            vis[v]=1;
            cnt++;
            q.push(v);
        }
    }

    return cnt;
}



int main()
{
    int n,m;
    cin>>n>>m;
    map<pii,int>mp;
    set<int>s;
    int mx=0;
    if(m==0) mx=1;
    while(m--)
    {
        int a,b;
        cin>>a>>b;
        if(mp[{a,b}] || mp[{b,a}]  ) continue;
        g[a].pb(b);
        g[b].pb(a);
        mp[{a,b}]++;
        mp[{b,a}]++;
        s.insert(a);
        s.insert(b);
    }


    for(auto u:s)
    {
        if(vis[u]==0)
        {
            mx=max(mx,bfs(u));
        }
    }



    cout<<mx<<endl;



	return 0;
}