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 numpy

class UnionFind:
    def __init__(self, N):
        self.N = N

        # the parent of all node is itself
        # self.parent = list(range(N))
        self.parent = [-1] * N

    def root(self, i):
        if self.parent[i] < 0:
            return i

        r = self.root(self.parent[i])
        self.parent[i] = r
        return r

    def unite(self, i, j):
        # print('root {}, {} = '.format(i, j), end='')
        i = self.root(i)
        j = self.root(j)
        # print(i, j)

        if i == j:
            return

        self.parent[i] += self.parent[j]
        self.parent[j] = i
        # print(self.parent)

    def same(self, i, j):
        return self.root(i) == self.root(j)

    def size(self, i):
        return -self.parent[self.root(i)]

N, M = map(int, input().split())
R = numpy.array([[int(r) for r in input().split()] for _ in range(M)]) - 1

forest = UnionFind(N)
for p1, p2 in R:
    forest.unite(p1, p2)

s = [forest.size(i) for i in range(N)]
print(max(s))