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(2)
#include<bits/stdc++.h>
using namespace std;
const int man = 2e5+10;
#define IOS ios::sync_with_stdio(0)
#define ull unsigned ll
#define uint unsigned
#define pai pair<int,int>
#define pal pair<ll,ll>
#define IT iterator
#define pb push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);++i)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);--i)
#define endl '\n'
#define ll long long
const ll mod = 1e9+7;
map<pair<int,int>,int>vis;
vector<int>sp[man];
int vi[man];

void dfs(int u,int f,int &num){
    ++num;
    vi[u] = 1;
    for(auto v:sp[u]){
        if(v==f||vi[v])continue;
        dfs(v,u,num);
    }
}

signed main() {
    #ifndef ONLINE_JUDGE
        //freopen("in.txt", "r", stdin);
        //freopen("out.txt","w",stdout);
    #endif
    int n,m;cin >> n>>m;
    if(!m){
        cout<<1<<endl;
        return 0;
    }
    for(int i = 1;i <= m;++i){
        int u,v;scanf("%d%d",&u,&v);
        if(vis.count({u,v}))continue;
        vis[{u,v}] = vis[{v,u}] = 1;
        sp[u].push_back(v);
        sp[v].push_back(u);
    }
    int maxx = 0;
    for(int i = 1;i <= n;++i){
        if(!vi[i]){
            int num = 0;
            dfs(i,i,num);
            maxx = max(maxx,num);
        }
    }
    cout<<maxx<<endl;
    return 0;
}