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;
#define pb push_back
#define F first
#define S second
//#define biGinf 9e+18
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define inpv(v) for(auto &x:v) cin>>x
#define otpv(v) for(auto &x:v) cout<<x<<nwl
#define otpv2(v) for(auto &x:v) cout<<x<<" "
#define CASES int tt;cin>>tt; while(tt--)
#define for0(i,n) for(int i=0; i<n; i++)
#define for1(i,n) for(int i=1; i<=n; i++)
#define forr(i,r,n) for(int i=r; i<=n; i++)
#define forj(j,n) for(int j=n; j>=0; j--)
#define forx(x,v) for(auto &x:v)
#define sz(v) (int)(v).size()
#define rz(n) resize(n)
#define vec(x) vector < x >
using int64 = long long;
using vi = vector < int >;
using vi64 = vector < int64 >;
using mii = map <int , int >;
using vvi = vector < vi >;
using pii = pair < int , int >;
using vpii = vector < pii >;
const int inf = 2e9 + 7;
const char nwl = '\n';
const int mxx = 1e6 + 3;
int64 sumV(vi &v) {
	int64 s = 0;
	for (int x : v) {
		s += (int64)x;
	}
	return s;
}
int gcd(int a, int b) {
	if (!b) return a;
	return gcd(b, a % b);
}
//const int64 biGinf = 9e+18;
//const string yo = "Yes\n", no = "No\n";
int ans;
void dfs(vi v[], vec(bool) &vis, int i, int cnt = 1) {
	vis[i] = 1;
	ans++;
	forx(x, v[i]) {
		if (!vis[x]) {
			dfs(v, vis, x, cnt + 1);
		}
	}
}


int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	//ur code
	int n, m;
	cin >> n >> m;
	vi v[n + 1];
	for1(i, m) {
		int x, y;
		cin >> x >> y;
		v[x].pb(y);
		v[y].pb(x);
	}
	int mx = 0;
	vec(bool) vis(n + 1);
	for1(i, n) {
		ans = 0;
		dfs(v, vis, i);
		mx = max(mx, ans);
	}
	cout << mx;


	return 0;
}