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 <set>
#include <deque>
using namespace std;
using ll = int64_t;

int main() {
  ll n,m;
  cin >> n >> m;
  vector<set<ll>> E(n, set<ll>{});
  for(ll i=0; i<m; i++) {
    ll a,b;
    cin >> a >> b;
    a--; b--;
    E[a].insert(b);
    E[b].insert(a);
  }
  vector<int> SEEN(n, false);
  ll ans = 0;
  for(ll i=0; i<n; i++) {
    if(!SEEN[i]) {
      ll size = 0;
      deque<ll> Q;
      Q.push_back(i);
      while(!Q.empty()) {
        ll x = Q.front(); Q.pop_front();
        if(SEEN[x]) { continue; }
        SEEN[x] = true;
        size++;
        for(ll y : E[x]) {
          Q.push_back(y);
        }
      }
      ans = max(ans, size);
    }
  }
  cout << ans << endl;
}