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>
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;


#include "boost/multiprecision/cpp_int.hpp"
#include "boost/multiprecision/cpp_dec_float.hpp"
namespace mp = boost::multiprecision;
typedef mp::cpp_int LL;
typedef mp::number<mp::cpp_dec_float<1024>> DD;// 仮数部が1024ビットの浮動小数点数型(TLEしたら小さくする)

 
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define rrep(i,r,l) for(ll i=r;i>=l;i--)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-12;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"  //インタラクティブで消す！！！！！！！！！！！！！！！！！！！！！
#define SORT(v) sort(v.begin(),v.end())
#define ERASE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define POSL(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define POSU(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
template<class T,class S>
inline bool chmax(T &a, S b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T,class S>
inline bool chmin(T &a, S b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

void mod_print(ll k){
    ll P=50000;
    rep(y,1,P){
        ll x=mod(y*k);
        
        if(abs(x)<=P||x+P>=i_7){
            if(x+P>=i_7){
                x-=i_7;
            }
            cout<<x<<"/"<<y<<endl;
            //cout<<setprecision(5)<<(dd)x/(dd)y;
            return;
        }
    }
    cout<<"nun"<<endl;
}
#define all(c) c.begin(),c.end()
typedef vector<ll> vl;
//////////////////////////


class UnionFind {
public:
    vector < ll > par; // 各元の親を表す配列
    vector < ll > siz; // 素集合のサイズを表す配列(1 で初期化)
    // Constructor
    UnionFind(ll sz_): par(sz_), siz(sz_, 1) {
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1);
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }
    // Member Function
    // Find
    ll root(ll x) { // 根の検索
        while (par[x] != x) {
            x = par[x] = par[par[x]]; // x の親の親を x の親とする
        }
        return x;
    }
    // Union(Unite, Merge)
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        // merge technique（データ構造をマージするテク．小を大にくっつける）
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }
    bool issame(ll x, ll y) { // 連結判定
        return root(x) == root(y);
    }
    ll size(ll x) { // 素集合のサイズ
        return siz[root(x)];
    }
};
/////////////////////////////////////
//UnionFind uf(N);


int main(){fastio
    ll n,m;cin>>n>>m;
    UnionFind uf(n+1);
    while(m--){
        ll a,b;cin>>a>>b;
        uf.merge(a,b);
    }
    ll ans=0;
    rep(i,1,n){
        chmax(ans,uf.size(i));
    }
    cout<<ans<<endl;
    
    return 0;
}