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)

import sys
n, m = map(int, input().split())
info = [tuple(map(int, s.split())) for s in sys.stdin.readlines()]

pars = [-1]*n

def get_root(x):
    if pars[x] < 0:
        return x
    else:
        pars[x] = get_root(pars[x])
        return pars[x]

def combine(x, y):
    if x == y:
        return
    r_x = get_root(x)
    r_y = get_root(y)
    if r_x > r_y:
        r_x, r_y = r_y, r_x
    if r_x != r_y:
        pars[r_x] += pars[r_y] # 足して経路長を求めるのに使う
        pars[r_y] = r_x

def size(x):
    return -pars[get_root(x)]

for x, y in info:
    combine(x, y)

print(-min(pars))