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 defaultdict
from collections import deque
from collections import Counter
import math

def readInt():
	return int(input())
def readInts():
	return list(map(int, input().split()))
def readChar():
	return input()
def readChars():
	return input().split()

n,m = readInts()
root = [i for i in range(n+1)]
ab = []
for i in range(m):
	a,b = sorted(readInts())
	root[b] = str(a)


for i in range(len(root)):
	while 1:
		if type(root[i]) is str:
			n = int(root[i])
			root[i] = root[n]
		else:
			break

print(Counter(root).most_common(1)[0][1])