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"
using namespace std;

#define sz(x) (int)(x.size())
#define FOR(i,L,R) for(int i = L; i <= R; i++)
#define int long long

const int N = 2e5 + 2, inf = 1e18;

struct DSU
{
    int connected;
    int par[N], sz[N];
 
    void init(int n) 
    {
        for(int i=1;i<=n;i++)
        {
            par[i]=i;
            sz[i]=1;
        }
        connected=n;
    }
 
    int getPar(int k)
    {
        while(k!=par[k])
        {
            par[k]=par[par[k]];
            k=par[k];
        }
        return k;
    }
 
    int getSize(int k)
    {
        return sz[getPar(k)];
    }

    bool same(int u , int v)
    {
        return (getPar(u) == getPar(v));
    }
 
    bool unite(int u, int v)
    {
        int par1=getPar(u), par2=getPar(v);
 
        if(par1==par2)
            return false;
 
        connected--;
 
        if(sz[par1]>sz[par2])
            swap(par1, par2);
 
        sz[par2]+=sz[par1];
        sz[par1]=0;
        par[par1]=par[par2];
        return true;
    }
};

int32_t main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int n, m;
    cin >> n >> m;
    DSU dsu;
    dsu.init(n);

    while(m--) {
        int u, v;
        cin >> u >> v;
        dsu.unite(u, v);
    }  

    int ans = 0;

    FOR(i,1,n) {
        ans = max(ans, dsu.getSize(i));
    }
    cout << ans;
}