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 lli   long long int
#define pb push_back
#define rep(i,s,n) for(lli i=s;i<n;i++)
#define irep(i,n,s) for(lli i=n-1;i>=s;i--)
#define mp make_pair
#define mod 1000000007
using namespace std;

vector<int>graph[200005];


void dfs(int src, bool vis[], int &count)
{
    vis[src] = true;
    count++;
    for (int i = 0; i < graph[src].size(); i++)
    {
        if (!vis[graph[src][i]])
        {

            dfs(graph[src][i], vis, count);
        }
    }

}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio;
    cin.tie(NULL);
    int i, j, k, m, n;
    // cin >> t;


    cin >> n >> m;


    for (int i = 0; i < m; i++)
    {
        int u, v;
        cin >> u >> v;
        graph[u].pb(v);
        graph[v].pb(u);
    }

    bool vis[n + 1] = {false};
    int ans = 0;

    for (int i = 1; i <= n; i++)
    {
        if (!vis[i])
        {
            int count = 0;
            dfs(i, vis, count);

            ans = max(ans, count);
        }
    }
    cout << ans ;






}