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<cstdio>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<math.h>
#include<tuple>
#include<string.h>
#include <numeric> 
#include<unordered_map>
using namespace std; 
#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define pp(...) "\033[94m [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "]\033[0m"
#define pp1(...) "\033[1;47;35m" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "]\033[0m"
#define dg(x) debug() << pp1(x)
#define dg1(x) debug() << "[line : " << __LINE__ <<"]"<< pp(x)
#define dg2(x,y) debug() << "[line : " << __LINE__ <<"]"<< pp(x) pp(y)
#define dg3(x,y,z) debug() << "[line : " << __LINE__ <<"]"<< pp(x) pp(y) pp(z)
#define dg4(w,x,y,z) debug() << "[line : " << __LINE__ <<"]"<< pp(w) pp(x) pp(y) pp(z)
#define int long long
#define all(x) x.begin() , x.end()
#define clr(x , y) memset(x , y , sizeof x);
#define sz(x) (int)x.size()
template<typename T> void db(T x ){ debug() <<">>> " <<  x; }
template<typename T> void db(T x , T y){ debug() <<">>> " <<  x <<" , "<< y; }
template<typename T> void db(T x , T y , T z){ debug() <<">>> "<< x <<" , "<< y <<" , "<< z; }
template<typename T> void db(T x , T y , T z , T a){ debug() <<">>> "<< x <<" , " << y <<" , "<< z <<" , "<< a; }
const int xx = 2e5;
const int Mod = 1e9 + 7;
const int inf = 1e18;
const int Maxn = 20010;
int cnt = 0;
vector<vector<int>> g;
vector<int> vis;

void dfs(int v) {
  if(vis[v]) return ;
  vis[v] = 1;
  cnt++;
  for(auto i: g[v]){
    if(!vis[i]) {
      dfs(i);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n , m;
  cin >> n >> m;
  g = vector<vector<int>>(n);
  vis = vector<int>(n, 0);
  for(int i = 0; i < m; i++) {
    int u , v;
    cin >> u >> v;
    u--,v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int ans = 0;
  for(int i = 0; i < n; i++) {
    if(!vis[i]) {
      cnt = 0;
      dfs(i);
      ans = max(ans , cnt);
    }
  }
  cout << ans << '\n';
}