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: Accepted)

import bisect,collections,copy,itertools,math,string
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def main():
    n,m = LI()
    route = [[] for _ in range(n+1)]
    for _ in range(m):
        a,b = LI()
        route[a].append(b)
        route[b].append(a)
    explored = set()

    sstock = collections.deque([i for i in range(1,n+1)])
    ans = 0
    while len(sstock)!=0:
        nnext = sstock.pop()
        if nnext not in explored:
            cnt = 0
            stock = collections.deque([nnext])
            explored.add(nnext)
            while len(stock) != 0:
                next = stock.pop()
                cnt += 1   
                for i in route[next]:
                    if i not in explored:
                        stock.append(i)
                        explored.add(i)
            ans = max(ans,cnt)
    print(ans)
main()