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;
typedef long long ll;
typedef long double ld;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for(int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
typedef pair<pi, int> pii;
typedef pair<pi, pi> pipi;
#define f first
#define s second
typedef vector<int> vi;
typedef vector<pi> vpi;
typedef vector<pii> vpii;
#define pb push_back
#define pf push_front
#define all(v) v.begin(), v.end()
#define disc(v) sort(all(v)); v.resize(unique(all(v)) - v.begin());
#define INF (int) 1e9 + 100
#define LLINF (ll) 1e18
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define sandybridge __attribute__((optimize("Ofast"), target("arch=sandybridge")))
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline ll rand(ll x, ll y) { ++y; return (rng() % (y-x)) + x; } //inclusivesss

int n, m, a, b, p[200005], sz[200005], ans;

int fs(int x) {
	if (p[x] == x) return x;
	return p[x] = fs(p[x]);
}

int main() {
	fastio; cin >> n >> m;
	FOR(i, 1, n) p[i] = i, sz[i] = 1;
	FOR(i, 1, m) {
		cin >> a >> b;
		if (fs(a) == fs(b)) continue;
		sz[fs(b)] += sz[fs(a)];
		p[fs(a)] = fs(b);
	}
	FOR(i, 1, n) ans = max(ans, sz[i]);
	cout << ans;
}