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 <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <set>

#define ll long long
#define REP(i,n) for(int i = 0; i < n; i++)
#define PI 3.141592653589
#define MOD 1000000007

using namespace std;

template<typename T> T gcd(T a, T b){return (a % b)? gcd(b, a % b) : b;}
template<typename T> T lcm(T a, T b){return a * b / gcd(a, b);}

class UnionFind{
    public:
    vector<int> parent;

    UnionFind(int N = 0) : parent(N, -1){}

    int root(int x){
        if(parent[x] < 0) return x;
        return parent[x] = root(parent[x]);
    }

    void unite(int x, int y){
        x = root(x);
        y = root(y);

        if(x == y) return;
        if(parent[x] > parent[y]) swap(x,y);

        parent[x] += parent[y];
        parent[y] = x;
    }
    
    bool same(int x, int y){ return (root(x) == root(y));}

    int size(int x){ return -parent[root(x)];}
};

int main(void){
    int n, m;
    cin >> n >> m;

    UnionFind uf(n);
    REP(i,m){
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        uf.unite(a,b);
    }

    int ans = -1;
    REP(i,n){
        ans = max(ans, uf.size(i));
    }
    cout << ans << endl;
    
    return 0;
}