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)

N,M = tuple(map(int,input().split()))
matchlist = [tuple(map(lambda x:int(x)-1,input().split())) for i in range(M)]
#最大のグループの大きさを出す
friendlist = [0 for i in range(N)]
gplist = [[]]
def match(ab):
    a,b=ab
    gpa=friendlist[a]
    gpb=friendlist[b]
    if bool(gpa) and gpa==gpb:
        return
    elif not bool(gpa) and not bool(gpb):
        friendlist[a]=len(gplist)
        friendlist[b]=len(gplist)
        gplist.append([a,b])
    elif bool(gpa) and not bool(gpb):
        friendlist[b] = gpa
        gplist[gpa].append(b)
    elif not bool(gpa) and bool(gpb):
        friendlist[a] = gpb
        gplist[gpb].append(a)
    else:
        for i in gplist[gpb]:
            friendlist[i] = gpa
        gplist[gpa].extend(gplist[gpb])
        gplist.remove(gplist[gpb])
# print(gplist)
for i in range(M):
    match(matchlist[i])
    # print(gplist)
    # print(friendlist)
print(max([len(item) for item in gplist]))