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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <numeric>
#include <string>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <bitset>
#include <iomanip> 

using ll = long long;
using ull = unsigned long long;
using namespace std;

#define FALSE printf("false\n");
#define TRUE printf("true\n");
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i = 0;(i) < ((int)(n));++(i))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )
#define test(a) cout<<"line:"<<__LINE__<<"["<<(#a)<<": "<<(a)<<"]"<<endl

const int INF = 1e9+7;
const ll INFL = 9*1e18;

// const int dx[4] = { 1, 0, -1, 0 };
// const int dy[4] = { 0, 1, 0, -1 };

const int dx[8] = {1, 1, 0,-1,-1,-1, 0, 1};
const int dy[8] = {0, 1, 1, 1, 0,-1,-1,-1};

ll inline digit(ll num){int tmp = 0;while(num){tmp++;num/=10;}return tmp;} // 桁数
template<typename T>inline void print(T&& x){cout<<x<<endl;}// 改行付き出力
template<typename T>inline T SUM(vector<T> vec){return accumulate(all(vec),(T)0);} // vectorの中身を全部足す
template<typename T>inline T digitSum(T num){T sum = 0;while(num){sum+=num%10;num/=10;}return sum;} // 各桁の和
template<typename T>inline T gcd(T a,T b){if(b == 0)return a;return gcd(b,a%b);} // 最大公約数
template<typename T>inline T lcm(T a, T b){T g = gcd(a,b);return a/g*b;} // 最小公倍数
template<typename T>inline T power(T a,T b){T tmp=1;rep(i,b){tmp*=a;}return tmp;} // 累乗
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int ctoi(char c){return c-'0';}


struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<int> siz;
    UnionFind(int N) : par(N),siz(N,1) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        if(siz[rx] < siz[ry])swap(rx,ry);
        siz[rx] += siz[ry];
        par[ry] = rx; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    int size(int x){
      return siz[root(x)];
    }
};

// TODO: write
int main() {
  int n,m;cin>>n>>m;
  auto tree = UnionFind(n);
  set<int> s;
  rep(i,m){
    int a,b;cin>>a>>b;
    tree.unite(a,b);
    s.emplace(a);
    s.emplace(b);
  }
  int ans = 0;
  rep(i,n){
    if(s.count(i+1))
    chmax(ans,tree.size(i+1));
  }
  print(ans);
  return 0;
}