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 <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <utility>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iterator>
#include <algorithm>
using namespace std;


const int N = 2e5 + 5;
vector<int> g[N];
bool vis[N];

void dfs(int u, int& sz)
{
    vis[u] = true;
    ++sz;
    for (int v : g[u])
    {
        if (!vis[v])
            dfs(v, sz);
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    map<pair<int, int>, int> edge;
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        int u, v;
        cin >> u >> v;
        if (edge.count({u, v}) == 0)
        {
            g[u].push_back(v);
            g[v].push_back(u);
            edge[ {u, v}] = 1;
            edge[ {v, u}] = 1;
        }
    }
    int ans = 0;
    memset(vis, false, sizeof vis);
    for (int i = 0; i < n; i++)
    {
        if (!vis[i])
        {
            int c = 0;
            dfs(i, c);
            //cout << i << " " << c << endl;
            ans = max(ans, c);
        }
    }
    cout << ans;
    return 0;
}