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 all(v) v.begin(), v.end()
#define x first
#define y second
#define PQ priority_queue
#define pb push_back
#define fu(i,a,j) for(int i=a;i<j;i++)
#define fd(i,a,j) for(int i=a;i>=j;i--)
#define SYNC ios::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL)
#define MOD 1000000007 
#define MOD2 1000000021
#define MAXN 2000000
#define INF 1e9
using namespace std;

typedef long long ll;
typedef long double ld;
typedef double db;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<string> vs;

//ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll m_pow(ll a, ll b, ll M = MOD) { ll res = 1; for (; b; b >>= 1, a = (a * a) % M)if (b & 1)res = (res * a) % M; return res; }
ll gcd(ll a, ll b) { if (a < b) swap(a, b); for (; b; a %= b, swap(a, b)); return a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); };
db dis(pii a, pii b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); }
int dr[] = { 0, 1, 0, -1, 1,1,-1,-1 };
int dc[] = { 1, 0, -1, 0, 1,-1,1,-1 };

int n, m, k, t;

string ansok[2] = { "NO","YES" };

pii a[200005];
int par[200005];
int sz[200005];
int Find(int u) {
	if (par[u] < 0) { return u; }
	return par[u] = Find(par[u]);
}
bool Union(int u, int v) {
	u = Find(u);
	v = Find(v);
	if (u == v)
		return 0;
	par[v] = u;
	sz[u] += sz[v];
	return 1;
}
int main()
{
	SYNC;
	memset(par, -1, sizeof par);
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		sz[i] = 1;
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		Union(u,v);
	}
	int mx = 0;
	for (int i = 1; i <= n; i++) {
		mx = max(mx, sz[Find(i)]);
	}
	cout << mx << '\n';
	return 0;
}