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>
const int MAX=2e5+10;
const int modn=1e9+7;
typedef long long ll ;
#define endl '\n'
#define cwk     freopen("D:\\workplace\\CLion\\in.in","r",stdin),freopen("D:\\workplace\\CLion\\out.out","w",stdout)
using namespace std;
int n,m,a,b;
int s[MAX];
int find_(int x){
    if(s[x]!=x) s[x]=find_(s[x]);
    return s[x];
}
void union_(int x,int y){
    x=find_(x);
    y=find_(y);
    if(x!=y)
        s[x]=s[y];
}
vector<int>v[MAX];
void solve(){
    for(int i=0;i<MAX;i++)
        s[i]=i;
    cin>>n>>m;
    while(m--){
        cin>>a>>b;
        union_(a,b);
    }
    int maxn=0;
    for(int i=1;i<=n;i++) {
        s[i]=find_(s[i]);
        v[s[i]].push_back(i);
        int p=v[s[i]].size();
        maxn=max(maxn,p);
    }
    cout<<maxn;

}
signed main() {
    //ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    //cwk;
    int _=1;
    //cin>>_;
    while(_--){
        solve();
    }
    return 0;
}