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 Python (3.8.2) (judge verdict: Accepted)

from collections import Counter
N,M = (int(T) for T in input().split())
def FindParent(X):
    if Parent[X]==X:
        return X
    else:
        Parent[X] = FindParent(Parent[X])
        return Parent[X]

def IsSameParent(X,Y):
    return FindParent(X)==FindParent(Y)

def UniteParent(X,Y):
    X = FindParent(X)
    Y = FindParent(Y)
    if X==Y:
        return 0
    if Rank[X]<Rank[Y]:
        Parent[X] = Y
    else:
        Parent[Y] = X
        if Rank[X]==Rank[Y]:
            Rank[X] += 1
Parent = [I for I in range(N+1)]
Rank = [0]*(N+1)
for TM in range(0,M):
    A,B = (int(T) for T in input().split())
    UniteParent(A,B)
for TN in range(1,N+1):
    FindParent(TN)
print(Counter(Parent[1:]).most_common()[0][1])