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>
using namespace std;

#define ll long long

ll P[1000000];

void init(ll N){
    for(int i = 1; i <= N; ++i) P[i] = i;
}

ll root(ll a){
    if(P[a] == a) return a;
    return (P[a] = root(P[a]));
}

bool is_same_set(int a, int b){
    return root(a) == root(b);
}

void unite(ll a, ll b){
    P[root(a)] = root(b);
}

int main(){
    ll N, M, ans = 0;
    cin >> N >> M;
    vector<ll> A(M+1), B(M+1), buf(N, 0);
    for(int i = 1; i <= M; ++i) cin >> A[i] >> B[i];
    init(N);
    for(int i = 1; i <= M; ++i){
        unite(A[i], B[i]);
    }
    for(int i = 1; i <= M; ++i){
        unite(A[i], B[i]);
    }
    for(int i = 1; i <= N;  ++i){
        ++buf[root(i)];
    }
    for(int i = 1; i <= N;  ++i){
        ans = max(ans, buf[i]);
    }
    cout << ans << endl;
    return 0;
}