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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from collections import deque
class Node:
    def __init__(self):
        self.friends = {}
        self.checked = False
    def add(self, b):
        self.friends[b] = True
N, M = [int(i) for i in input().split()]
P = [Node() for i in range(N)]
for i in range(M):
    A, B = [int(i)-1 for i in input().split()]
    P[A].add(B)
    P[B].add(A)

counter = 0
for i in range(N):
    if P[i].checked == False:
        d = deque([i])
        counter += 1
    else:
        continue
    while d:
        q = d.popleft()
        P[q].checked = True
        for key in P[q].friends.keys():
            if P[key].checked == False:
                d.append(key)
            for j in P[key].friends.keys():
                if P[key].checked == False:
                    d.append(j)
print(counter)