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: Accepted)

from copy import deepcopy

N, M = map(int, input().split())

pairs = [0] * M
for i in range(M):
    pairs[i] = list(map(int, input().split()))


# N = 5
# pairs = [[1,2], [3,4], [5,1]]


class UnionFind:
    def __init__(self, N):
        self.r = [-1] * N

    def root(self, x):
        if self.r[x] < 0:
            return x
        self.r[x] = self.root(self.r[x])
        return self.r[x]


    def unite(self, x, y):
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return False
        if self.r[x] > self.r[y]:
            x, y = y, x

        self.r[x] += self.r[y]
        self.r[y] = x
        return True

    def size(self, x):
        return -self.r[self.root(x)]

uf = UnionFind(N)
for pair in pairs:
    a = pair[0] - 1
    b = pair[1] - 1
    uf.unite(a, b)

ans = 0
for i in range(N):
    ans = max(ans, uf.size(i))

print(ans)