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: Wrong Answer)

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
#define forn(i,n) for(ll i=0;i < ll (n); i++)

int main()
{
    ll n,m;
    cin>>n>>m;
    ll a[n+1];
    forn(i,n+1) a[i]=-1;
    ll j,k;
    vector<pair<ll,ll> >v;
    map<ll, ll>mp;
    ll ans=1;
    forn(i,m)
    {
        cin>>j>>k;
        if(a[j]==-1 && a[k]==-1) a[j]=j, a[k]=j, mp[j]+=2, ans=max(mp[j], ans);
        else if(a[j]==a[k]) continue;
        else if(a[j]==-1) a[j]=a[k], mp[a[j]]++, ans=max(mp[a[j]], ans);
        else if(a[k]==-1) a[k]=a[j], mp[a[k]]++, ans=max(mp[a[k]], ans);
        else v.push_back({a[j], a[k]});
    }
    forn(i,v.size())
    {
        ll u1=v[i].first;
        ll u2=v[i].second;
        ans=max(ans, mp[u1]+mp[u2]);
    }
    cout<<ans<<endl;
    return 0;
}