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)

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>

using namespace std;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define ll long long int



int main(){
    int n,m;
    cin>>n>>m;
    vector<int> a(n,-1);
    vector<vector<int>> tree(n);
    for(int i=0;i<m;i++){
        int u,v;
        cin>>u>>v;
        u--,v--;
        tree[u].push_back(v);
        tree[v].push_back(u);
    }
    vector<bool> inc(n,false);
    function<void(int,int)> dfs =[&](int node,int cur){
        debug(node,cur);
        a[node]=cur;
        inc[node]=true;
        for(auto v:tree[node]){
            if(!inc[v])dfs(v,cur);
        }
    };
    int gr=1;
    for(int i=0;i<n;i++){
        if(!inc[i])dfs(i,gr++);
        
    }
    debug(a);
    vector<int> cnt(gr);
    for(int i=0;i<n;i++){
        cnt[a[i]-1]++;
    }
    debug(cnt);
    cout<<*max_element(cnt.begin(),cnt.end())<<"\n";
    return 0;
}