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 rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const double pi = acos(-1.0);
const double EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

struct Union_Find_Tree{
    vector<int> par, rank, num;
    
    Union_Find_Tree(int N){
        par.assign(N, -1);
        rank.assign(N, 0);
        num.assign(N, 1);
    }
    
    int root(int x){
        if(par[x] < 0 || par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    
    bool unite(int x, int y){
        x = root(x), y = root(y);
        if(x == y) return false;
        if(rank[x] < rank[y]) swap(x, y);
        if(rank[x] == rank[y]) rank[x]++;
        par[y] = x, num[x] += num[y];
        return true;
    }
    
    int size(int x) {return num[root(x)];}
    
    bool same(int x, int y){
        return root(x) == root(y);
    }
    
    void clear(){
        fill(par.begin(), par.end(), -1);
        fill(rank.begin(), rank.end(), 0);
        fill(num.begin(), num.end(), 1);
    }
};

int main(){
    int N, M;
    cin >> N >> M;
    Union_Find_Tree uf(N);
    rep(i, M){
        int u, v; cin >> u >> v; u--, v--;
        uf.unite(u, v);
    }
    int ans = 0;
    rep(i, N) chmax(ans, uf.size(i));
    cout << ans << endl;
}