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>

typedef long long ll;
using namespace std;

#define _overload(_1, _2, _3, name, ...) name
#define _rep(i, n) for(ll i = 0; i < (ll)n; i++)
#define _repi(i, s, n) for(ll i = (ll)(s); i < (ll)n; i++)
#define REP(...) _overload(__VA_ARGS__, _repi, _rep,)(__VA_ARGS__)
#define all(v) (v).begin(),(v).end()
#define UNIQUE(v) (v).erase( unique( all(v) ), (v).end() );
#define MOD 1000000007

int main(){
	ll n, m;	cin >> n >> m;
	vector<pair<ll ,ll>> f(m*2);
	REP(i, m){
		cin >> f[i].first >> f[i].second;
		f[i].first--;
		f[i].second--;
	}
	vector<ll> group[n];
	ll ans = 0;

	REP(i, m){
		f[i+m].first = f[i].second;
		f[i+m].second = f[i].first;
	}
	sort( all(f) );
	UNIQUE( f );

	REP(i, n)	group[i].push_back(i);

	REP(i, f.size()){
		group[ f[i].first ].push_back( f[i].second );
	}

	REP(i, n){
		ans = max(ans, (ll)group[i].size());
	}

	cout << ans << endl;

	return 0;
}