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(i = 0;i < n;++i)
#define all(v) v.begin(), v.end()
using ll = long long;

struct UnionFind
{
    int i;
    vector<int> par;
    vector<int> size;
    UnionFind(int n): par(n),size(n,1){
        for(i = 0;i < n;++i){
            par.at(i) = i;
        }
    }
    
    int root(int x){
        if(par.at(x) == x){
            return x;
        }else{
            return par.at(x) = root(par.at(x));
        }
    }
    
    void unite(int x, int y){
        int rx = root(x);
        int ry = root(y);
        if(rx == ry){
            return;
        }else{
            if(size.at(rx) < size.at(ry)){
                swap(rx,ry);
            }
            par.at(ry) = rx;
            size.at(rx) += size.at(ry);
            return;
        }
    }
    
    bool same(int x, int y){
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
    
    int get_size(int x){
        return size.at(root(x));
    }
};

int main()
{
    ll i,j;
    ll n,m;
    cin >> n >> m;
    UnionFind uf(n);
    for(i = 0;i < m;++i){
        ll a,b;
        cin >> a >> b;
        --a;    --b;
        uf.unite(a,b);
    }
    ll ans = 0;
    for(i = 0;i < n;++i){
        ll tmp = uf.get_size(i);
        ans = max(ans, tmp);
    }
    cout << ans << endl;

    return 0;
}