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: Accepted)

#n = int(input())
import random
import sys
sys.setrecursionlimit(10 ** 7)
n, m = map(int, input().split())
#n=200000
#m=200000
#al = list(map(int, input().split()))
#l = [list(map(int,input().split())) for i in range(n)]
adjl = [[] for _ in range(n+1)]
NIL = -1
appear = {}
isalone = {}
for i in range(m):
    a, b = map(int, input().split())
    #a=random.randrange(1,n+1)
    #b=random.randrange(1,n+1)
    #while a==b:
    #    b=random.randrange(1,n+1)
    #print(a,b)
    if appear.get((a, b), False) or appear.get((a, b), False):
        continue
    appear[(a, b)] = True
    appear[(b, a)] = True
    isalone[a] = False
    isalone[b] = False
    adjl[a].append(b)
    adjl[b].append(a)

color = [NIL for i in range(n+1)]
color_id = 0


def dfs(u, color_id):
    global color
    color[u] = color_id
    for v in adjl[u]:
        if color[v] == NIL:  # 未到達なら再起呼び出し
            color[v] = color_id
            dfs(v, color_id)


for u in range(1, n+1):
    if color[u] == NIL:
        color_id += 1
        dfs(u, color_id)

group = {}
for i in range(1, n+1):
    if isalone.get(i, True):
        continue
    group[color[i]] = group.get(color[i], 0)+1

mx = 1
for k, v in group.items():
    mx = max(mx, v)

print(mx)