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 rep(i,n)    for(long long i = 0; i < (n) ; ++i)
#define orep(i,n)   for(long long i = 1; i <= (n); ++i)   // one rep
#define drep(i,n)   for(long long i = (n)-1; i >= 0; --i) // down rep
#define srep(i,s,t) for(long long i = (s); i < (t); ++i)  // set rep
#define rng(x)  (x).begin(),(x).end()

  // range
#define rrng(a) (a).rbegin(),(a).rend() // reverse range
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define ru(x,y) (((x)+(y)-1)/(y)) // round up
#define fi first
#define se second
#define eb emplace_back
#define fcout cout << fixed << setprecision(15)


using namespace std;
using ll  = long long;
using P   = pair<int,int>;
using vi  = vector<int>;
using vvi = vector<vi>;
using vp  = vector<P>;
using vl  = vector<ll>;

template<typename T> void Yes(T flag) {
  cout << (flag ? "Yes" : "No") << endl;
}

template<typename T> void pv(vector<T> vec) {
  cout << "[";
  for(auto& v : vec) { cout << v << ","; }
  cout << "]" << endl;
}

template<>
void pv(vector<P> vec) {
  cout << "[";
  for(auto& v : vec) {
    cout << "<" << v.fi << ","<< v.se << ">" << ",";
  }
  cout << "]" << endl;
}

template<typename T> void pv2(vector<vector<T>> vec) {
  for(auto& v : vec) pv(v);
}

int n;
vvi maze;
vi seen;
int mxlength;
vi p;
int tmp;

int dfs(int x, int length) {
  int sz = maze.at(x).size();
  mxlength++;
  seen.at(x) = 1;
  rep(i,sz) {
    int nx = maze.at(x).at(i);
    if(!seen.at(nx)) {
      dfs(nx,length+1);
    }
  }
  return length;
}

int main() {
  int n,m;cin >> n >> m;

  maze = vvi(n,vi());
  seen = vi(n,0);
  rep(i,m) {
    int a,b;cin >> a >> b;
    a--;b--;
    maze.at(a).eb(b);
    maze.at(b).eb(a);
  }


  int ans = 0;
  rep(i,n) {
    int length=0;
    mxlength = 0;
    if(!seen.at(i)) {
      dfs(i,length);
      ans = max(ans,mxlength);
    }
  }

  cout << ans << endl;


  return 0;


  return 0;
}