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 ll long long
#define ld long double
#define F first
#define S second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define ppb pop_back
#define vll vector<ll>
#define vi vector<int>
#define endl "\n"

const ll INF = 1000000007;


const int max_size = 200005;
int parent[max_size];
int Size[max_size];

int root(int x){
	while(x!=parent[x]){
		parent[x] = parent[parent[x]];
		x = parent[x];
	}
	return x;
}

void sunion(int x,int y){
	int rx = root(x);
	int ry = root(y);
	if(rx != ry){
		if(Size[rx] > Size[ry]){
			parent[ry] = parent[rx];
			Size[rx] += Size[ry];
		}
		else{
			parent[rx] = parent[ry];
			Size[ry] += Size[rx];
		}
	}
}


int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int tc=1;
	//cin >> tc;
	while(tc--){
		int n;
		cin >> n;
		int m;
		cin >> m;
		int u,v;
		for (int i=1;i<=n;i++) {
			parent[i] = i;
			Size[i] = 1;
		}
		for (int i=0;i<m;i++) {
			cin >> u >> v;
			sunion(u,v);
		}
		int maxSize = 0;
		for (int i=1;i<=n;i++) {
			int r = root(i);
			maxSize = max(maxSize,Size[r]);
		}
		cout << maxSize << endl;
	}
	return 0;
}