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: Time Limit Exceeded)

#include<iostream>
#include<cstdio>
#include <stdio.h>
#include<algorithm>
#include<cstring>
#include <string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<vector>
#include<bits/stdc++.h>
#include <set>
#define ll   long long
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#define inf 0x3f3f3f3f
#define pi 3.1415926535898
using namespace std;
const int N=2e5+10;
const int mod=1e9+7;
ll f[N+1];
ll b[N+1];
ll finds(int x)
{
    if(x!=f[x])
    {

        return  finds(f[x]);
    }
    return x;
}
void uni(int x,int y)
{

    int x1=finds(x);
    int y1=finds(y);
    int a=min(x1,y1);
    int b=max(x1,y1);
    f[b]=f[a];

}
int main()
{
    for(int i=1; i<=N; i++)
    {
        f[i]=i;
    }
    int n,m;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        int x,y;
        cin>>x>>y;
        uni(x,y);
    }
    ll maxs=-99999;
    for(int i=1; i<=n; i++)
    {
        b[finds(i)]++;
    }
    for(int i=1; i<=n; i++)
    {
        maxs=max(maxs,b[i]);
    }
    cout<<maxs<<endl;
    return 0;
}