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)

def main():
    n,m = [int(v) for v in input().split()]
    groups = []
    g = [-1]*(n+1)
    for _ in range(m):
        a,b = [int(v) for v in input().split()]
        if g[a]!=-1 and g[b]!=-1:
            if g[a]==g[b]: continue
            gb = g[b]
            for no in groups[gb]:
                g[no] = g[a]
            groups[gb] = {}
        elif g[a]!=-1:
            g[b] = g[a]
            groups[g[a]].append(b)
        elif g[b]!=-1:
            g[a] = g[b]
            groups[g[b]].append(a)
        else:
            g[a] = g[b] = len(groups)
            groups.append([a,b])
    # print(groups, g)
    ans = max([len(v) for v in groups])
    print(ans)

main()