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>

using namespace std;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using vs = vector<string>;
using pll = pair<ll, ll>;
const ll oo = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c),end(c)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define FOR(i, a, b) for(ll i=(a); i<(b); i++)
#define FORD(i, a, b) for(ll i=ll(b)-1;i>=(a);i--)
#define TR(X) ({if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

#define MOD 1000000007

ll n, m;
vvl graph;

vb visited;

ll dfs(ll node) {
    if(visited[node]) return 0;
    visited[node] = true;
    ll output = 1;
    for(ll neigh : graph[node]) {
        output += dfs(neigh);
    }
    return output;
}

int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;
    graph = vvl(n);
    FOR(i, 0, m) {
        ll a, b; cin >> a >> b;
        a--; b--;
        graph[a].pb(b);
        graph[b].pb(a);
    }

    visited = vb(n, false);
    ll output = 0;
    FOR(i, 0, n) {
        if(visited[i]) continue;
        output = max(output, dfs(i));
    }

    cout << output << endl;
}