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 <iostream>
#include <vector>
#include <deque>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <list>
#include <unordered_set>
#include <tuple>
#include <cmath>
#include <limits>
#include <type_traits>
#include <iomanip>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <set>
#include <bitset>
#include <regex>
#include <random>
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
#define rep(i,n)for(ll i=0;i<n;++i)
#define exout(x) printf("%.10f\n", x)
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
const double pi = acos(-1.0);
const ll MOD = 1000000007;
const ll INF = 1e18;
const ll MAX_N = 201010;

//組み合わせの余りを求める
ll fac[MAX_N], finv[MAX_N], inv[MAX_N];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX_N; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(ll n, ll k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

ll lcm(ll x, ll y) {
	if (x == 0 || y == 0)return 0;
	return (x / gcd(x, y) * y);
}

ll dx[4] = { 0,0,-1,1 };
ll dy[4] = { -1,1,0,0 };
char notes[110][10];
bool flag[201010];
vector<vector<ll>>to;
ll res;

void dfs(ll x) {
	if (flag[x] == false) {
		res++;
		flag[x] = true;
		for (auto nx : to[x]) {
			if (flag[nx]==false) {
				dfs(nx);
			}
		}
	}
}

//long longしか使わない
//素数は1より大きい
//lower_boundは指定したkey以上の要素の一番左のイテレータをかえす
//upper_boundは指定したkeyより大きい要素の一番左のイテレータをかえす
int main() {
	ll n, m;
	cin >> n >> m;
	to.resize(n);
	rep(i, m) {
		ll a, b;
		cin >> a >> b;
		a--;
		b--;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	ll ans = 0;
	rep(i, n) {
		res = 0;
		if (flag[i] == false) {
			dfs(i);
		}
		ans = max(res, ans);
	}
	cout << ans << endl;
	return 0;
}