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: Wrong Answer)

#include <bits/stdc++.h>

int main(){
    using namespace std;
    unsigned long N, M;
    cin >> N >> M;
    vector<unsigned long> uf(N), sz(N, 1);
    iota(begin(uf), end(uf), 0UL);
    const auto& find = [&uf](auto i) -> unsigned long {
        while(uf[i] != i)i = uf[i] = uf[uf[i]];
        return i;
    };
    const auto& unite = [&uf, &sz, &find](auto i, auto j) -> unsigned long {
        i = find(i);
        j = find(j);
        if(i == j)return 0;
        if(sz[i] < sz[j])swap(i, j);
        sz[i] += sz[j];
        uf[j] = i;
        return sz[i];
    };
    unsigned long ans{0};
    for(unsigned long i{0}, a, b; i < M; ++i){
        cin >> a >> b;
        ans = max(ans, unite(--a, --b));
    }
    cout << ans << endl;
    return 0;
}