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: Wrong Answer)

#include    <bits/stdc++.h>
#include    <ext/pb_ds/assoc_container.hpp>
#include    <ext/pb_ds/trie_policy.hpp>
#include    <ext/rope>

#pragma     03
#pragma     GCC             optimize("Ofast")
#pragma     GCC             optimization("unroll-loops, no-stack-protector")
#pragma     GCC             target("sse,sse2,sse3,ssse3,sse4.1,sse4.2")

#define     ll              long long
#define     ld              long double
#define     fi              first
#define     se              second
#define     sz(x)           x.size()
#define     mp              make_pair
#define     pb              push_back
#define     eb              emplace_back
#define     all(x)          begin(x), end(x)
#define     rall(x)         rbegin(x), rend(x)
#define     rep(i, x, n)    for (ll i = x; i < n; i++)
#define     fastio()        ios_base::sync_with_stdio(NULL); cin.tie(0); cout.tie(0)
#define     int             long long
using       namespace std;
using       namespace __gnu_pbds;
using       namespace __gnu_cxx;

mt19937     rng(chrono::steady_clock::now().time_since_epoch().count());

typedef     vector<int>     vi;
typedef     vector<vi>      vvi;
typedef     pair<int, int>  pii;
typedef     vector<pii>     vii;
typedef     vector<ll>      vl;
typedef     vector<vl>      vvl;
typedef     pair<ll, ll>    pll;
typedef     vector<pll>     vll;

inline void EXECUTE_TIME() {cout << endl << setprecision(10) << fixed << "CPU Time: " << clock() / double(CLOCKS_PER_SEC) << " sec(s)\n";}

template <class T> inline T RD(T &a) {T x; cin >> x; return a = x;}
template <class T> inline T gcd(T a, T b) {if (b) return gcd(b, a % b); else return a;}
template <class T> inline T lcm(T a, T b) {return a / gcd(a,b) * b;}
template<typename T> using ordered_set=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef trie<string, null_type, trie_string_access_traits<>, pat_trie_tag, trie_prefix_search_node_update> pref_trie;

const int N = 2e5 + 1;

int n;

int par[N], setSize[N];
int ans = -1;

int find(int i) {
    while (i != par[i]) {
        i = par[i];
    }
    return par[i];
}

void join(int i, int j) {
    if ((i = find(i)) != (j = find(j))) {
        par[j] = par[i];
        setSize[i] += setSize[j];
        ans = max(ans, setSize[i]);
    }
}

signed main() {
//    freopen("bubbletea.inp", "r", stdin);
//    freopen("bubbletea.out", "w", stdout);
    fastio();

    int m; cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        par[i] = i;
        setSize[i] = 1;
    }
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        join(a, b);
    }

    cout << ans;

//    EXECUTE_TIME();
}