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 ll long long
#define ld long double
#define pii pair<ll, ll>
#define SZ(x) ((int)(x).size())
#define endl "\n"
#define vvi vector<vector<int> >
#define vvl vector<vector<long long> >
ll mod = 1e9 + 7;
//const double pi = acos(-1.0);
/*
ll kmp(string x)
{
    ll n= x.length();
    //cout<<x<<endl;
    vector<ll> lcp(x.length(), 0);
    for(ll i=1;i<n;i++)
    {
        ll j=lcp[i-1];
        while(j>0 && x[j]!=x[i])
            j=lcp[j-1];
        if(x[i]==x[j])
            j++;
        lcp[i]=j;
    }
    return lcp[n-1];
}

int N = 1000002;
int ar[100002];
void sieve()
{
    for(int i=1;i<=N;i++)
        ar[i] = 1;
    ar[0] = 0;
    for(int i=2;i*i<=N;i++)
    {
        if(ar[i] && i*i<=N)
        {
            for(int j = i*i; j<=N;j+=i)
                ar[j] = 0;
        }
    }
}

ll modex(ll a,ll b){
    if(b<=0){
        return 1;
    }
    if(b%2==0)
    {
        ll x = modex(a, b/2);
        return ((x*x)%mod);
    }
    else
        return ((a%mod * modex(a, b-1))%mod);
}

ll c[1010][1010];

ll combina(ll i, ll j){
        for (int i=0;i<=1005;i++)
        for (int j=0;j<=i;j++)
        {
            if (j==i||j==0)
                c[i][j]=1;
            else
            c[i][j]=c[i-1][j-1]+c[i-1][j];
            c[i][j]%=mod;
        }
        return c[i][j];
}
void find_fac(int x){
    vector<ll> di;
    for(ll i = 2; i*i <= x; i++)
    {
        if(x%i==0){
            di.push_back(i);
            while(x%i==0)
                x/=i;
        }
    }
    if(x > 1)
        di.push_back(x);
}
*/
ll n, m;
vector<vector<ll> > gr;
vector<ll> v;
ll c = 0;
void dfs(ll x){
    v[x] = 1;
    c++;
    for(auto i:gr[x])
        if(!v[i])
            dfs(i);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cout<<fixed;
    cout.precision(12);
    ll n, m;
    cin>>n>>m;
    gr.resize(n+1);
    for(ll i=0;i<m;i++){
        ll x, y;
        cin>>x>>y;
        gr[x].push_back(y);
        gr[y].push_back(x);
    }
    ll an = 0;
    v.resize(n+1);
    for(ll i=1;i<=n;i++){
        if(!v[i]){
            c = 0;
            dfs(i);
            an = max(an, c);
        }
    }
    cout<<an<<endl;
    return 0;
}