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>
#include "bits/stdc++.h"

using namespace std;

#define int long long

#ifndef bhupixb
    #define var(...)
    #define stl(...)
#endif

#define single_test

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

int dfs(int x) {
    vis[x] = 1;
    int cnt = 1;
    for (int adj: g[x]) {
        if (!vis[adj]) {
            cnt += dfs(adj);
        }
    }
    return cnt;
}

void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    int mx = 0;
    for (int i = 1; i <= n; ++i) {
        if (!vis[i]) {
            mx = max(mx, dfs(i) - 1);
        }
    }
    cout << mx + 1 << '\n';
}

signed main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    #ifndef single_test
        cin >> t;
    #endif
    for (int i = 1; i <= t; ++i) {
        // cout << "Case #" << i << ": ";
        solve();
    }

    return 0;
}