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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

struct UnionFind {
  vector<int> d;
  UnionFind(int n = 0) : d(n, -1) {}

  int find(int x) {
    if (d[x] < 0)
      return x;
    return d[x] = find(d[x]);
  }

  bool unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y)
      return false;
    if (d[x] > d[y])
      swap(x, y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }

  bool same(int x, int y) { return find(x) == find(y); }
  int size(int x) { return -d[find(x)]; }
};

int main() {
  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 = 0;
  rep(i, n) ans = max(ans, uf.size(i));
  cout << ans << endl;

  return 0;
}