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 copy
(N,M) = map(int,input().split())
l = []
if M == 0:
    print(1)

else:
    for i in range(M):
        l.append([int(x) for x in input().split()])

    mn = 0
    for i in range(M):
        for j in range(2):
            mn = max(mn,l[i][j])

    s = [0 for i in range(mn+100)]
    k = [0 for i in range(M)]
    cnt = 0
    for i in range(M):
        if s[l[i][0]-1] == 0 and s[l[i][1]-1] == 0:
            cnt += 1
            s[l[i][0]-1] = cnt
            s[l[i][1]-1] = cnt
            k[cnt-1] += 2
        elif s[l[i][0]-1] == 0 and s[l[i][1]-1] != 0:
            s[l[i][0]-1] = s[l[i][1]-1]
            k[s[l[i][1]-1]-1] += 1
        elif s[l[i][0]-1] != 0 and s[l[i][1]-1] == 0:
            s[l[i][1]-1] = s[l[i][0]-1]
            k[s[l[i][0]-1]-1] += 1
        elif s[l[i][0]-1] != s[l[i][1]-1]:
            t = copy.deepcopy(s)
            s = [t[l[i][1]-1] for x in t if x == t[l[i][0]-1]]
            k[t[l[i][1]-1]-1] += k[t[l[i][0]-1]-1]

    print(max(k))