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;
long long mod=1e9+7;
long long INF=1e7;
double EPS = 1e-12;
const long long lim = 3e5+100;
typedef long long int lld;
typedef long long ll;

#define rep(i,a,n) for(long long int i = (a); i <= (n); ++i)
#define mset(a, v) memset(a, v, sizeof(a))
#define Error(x...) { cout << "(" << #x << ")" << " = ( "; printIt(x); }
template <typename T1> void printIt(T1 t1) { cout << t1 << " )" << endl; }
template <typename T1, typename... T2>
void printIt(T1 t1, T2... t2) { cout << t1 << " , "; printIt(t2...); }
 
 
#define popcount __builtin_popcountll
 
ll powP(ll x, ll y) {
    ll ans = 1, po = x % mod;
    while(y > 0) {
        if(y & 1) ans = ans * po % mod;
        y >>= 1; po = po * po % mod;
    }
    return ans;
}

vector<ll> Ans, Adj[lim];
set<ll> S[lim];
ll Array[lim], Vis[lim];
pair<ll,ll> P[lim];

ll dfs(ll node,ll Vis[]) {
    ll countA=1;
    Vis[node]=1;
    for(auto i: S[node]) {
        if(!Vis[i]) countA+= dfs(i,Vis);
    }

    return countA;
}

int main(){

    ios::sync_with_stdio(0);
    cin.tie(0);

    #ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input.txt", "r", stdin);
    // for writing output to output.txt
    freopen("output.txt", "w", stdout);
    #endif
    
    lld T,i,p,j,l,e,r,b,c,k,m,q,a,d,w,x,y,u,v,z,t,curr,prev,sum=0,val,countA=0,indicator=0,pos,h,minA,maxA,type,n,ans=0,left,right;
    string A,B;
    A = "Yes", B="No";
    cin>>n>>m;
    rep(i,1,m) {
        cin>>a>>b;
        S[a].insert(b);
        S[b].insert(a);
    }

    rep(i,1,n) {
        if(!Vis[i]) ans = max(ans,dfs(i,Vis));
    }

    cout<<ans<<endl;
    return 0;
}