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: Wrong Answer)

from collections import Counter
class DSU:
    def __init__(self, N):
        self.p = [i for i in range(0, N + 1)]

    def find(self, x):
        if x == self.p[x]:
            return x
        return self.find(self.p[x])

    def union(self, x, y):
        xp = self.find(x)
        yp = self.find(y)
        if xp != yp:
            if xp < yp:
                xp, yp = yp, xp
            self.p[xp] = yp

N, C = list(map(int, input().split()))
dsu = DSU(N)
for _ in range(C):
    x, y = list(map(int, input().split()))
    dsu.union(x, y)
c = Counter(dsu.p)
for _, v in sorted(c.items(), key = lambda x: -1*x[1]):
    print(v)
    break