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 pb emplace_back
using namespace std;
using ll = long long;
#ifdef KEV
#define DE(i, e) cerr << #i << ' ' << i << e
void debug(auto L, auto R) { while (L < R) cerr << *L << " \n"[L+1==R], ++L; }
#else
#define DE(...) 0
void debug(...) {}
#endif
const int maxn = 300010;
int n, m, g[maxn], sz[maxn];
int find(int i) { return i == g[i] ? i : g[i] = find(g[i]); }
void merge(int a, int b) {
	if ((a = find(a)) == (b = find(b))) return;
	if (sz[a] < sz[b]) swap(a, b);
	sz[a] += sz[b];
	g[b] = a;
}
signed main() {
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m;
	for (int i = 1;i <= n;++i)
		g[i] = i, sz[i] = 1;
	int res = 0;
	while (m--) {
		int a, b;
		cin >> a >> b;
		merge(a, b);
	}
	for (int i = 1;i <= n;++i)
		res = max(res, sz[i]);
	cout << res << '\n';
}