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 <iostream>
#include <vector>
using namespace std;

int find(vector<int> &p, int x) {
     if (p[x] != x) {
         p[x] = find(p, p[x]);
     }
     return p[x];
}

void uni(vector<int> &p, vector<int>& r, vector<int>& s, int x, int y) {
    int px = find(p, x);
    int py = find(p, y);
    if (px == py) {
        return;
    }

     if (r[px] < r[py]) {
        p[px] = py;
        s[py]++;
     } else if (r[px] > r[py]) {
        p[py] = px;
        s[px]++;
     } else {
        p[py] = px;
        r[px]++;
        s[px]++;
     }
}

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> p(n), r(n, 0), s(n, 1);
    for (int i = 0; i < n; i++) {
        p[i] = i;
    }
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        uni(p, r, s, a, b);
    }
    int mm = 1;
    for (int i = 0; i < n; i++) {
        if (s[i] > mm) {
            mm = s[i];
        }
    }
    cout << mm << endl;
}