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: Runtime Error)

from collections import *
n,m = map(int,input().split())
par = [-1 for _ in range(n)]#par<0なら自身が親で要素の個数、>0なら子で親の位置示す

def root(i):
    if par[i] < 0:
        return i
    else:
        return root(par[i])

def size(a):
    return -par[root(a)]

def union(a,b):
    a = root(a)
    b = root(b)
    if a == b:#親が等しい
        return False
    if size(a) < size(b):#サイズが大きい方に繋げる
        a,b = b,a
    par[a] += par[b]
    par[b] = a
    return True

for i in range(m):
    a = [int(j)-1 for j in input().split()]#数字を一つ減らして足を揃える
    if root(a[0]) != root(a[1]):
        union(a[0],a[1])
par = [ i for i in par if i >= 0]
print(Counter(par).most_common()[0][1]+1)