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 rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll INF = 1LL << 60;

struct UnionFind {
    vector<int> p;
    vector<int> r;

    UnionFind(int n){
        p.resize(n, -1);
        r.resize(n, 1);
    }

    int find(int x){
        if(p[x] == -1) return x;
        else return p[x] = find(p[x]);
    }

    void unite(int x, int y){
        x = find(x);
        y = find(y);
        
        if(x == y) return;
        if(r[x] > r[y]) swap(x, y);
        if(r[x] == r[y]) ++r[y];

        p[x] = y;
    }
};

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;

    UnionFind uf(n);

    rep(i, m){
        int a, b;
        cin >> a >> b;
        --a;
        --b;

        uf.unite(a, b);
    }

    vector<int> cnt(n, 0);
    rep(i, n){
        ++cnt[uf.find(i)];
    }

    int res = 0;
    rep(i, n) res = max(res, cnt[i]);

    cout << res << endl;

    return 0;
}