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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from sys import stdin
from collections import Counter


# Union-Find木
class Union_Find():
    def __init__(self, N=1):
        self.par = [i for i in range(N)]  # 親ノード
        self.rank = [0] * N               # ランク

    # 経路圧縮
    def root(self, x):
        if self.par[x] == x:
            return x
        else:
            r = self.root(self.par[x])
            self.par[x] = r
            return r

    # xとyが同じグループに属するか判定
    def issame(self, x, y):
        return self.root(x) == self.root(y)

    # xとyの属するグループを1つにまとめる
    def merge(self, x, y):
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return False
        else:
            if self.rank[x] < self.rank[y]:
                x, y = y, x
            elif self.rank[x] == self.rank[y]:
                self.rank[x] += 1
            self.par[y] = x
            return True


def main():
    _in = [_.rstrip() for _ in stdin.readlines()]
    N, M = list(map(int, _in[0].split(' ')))  # type:list(int)
    A_B_arr = []
    for i in range(M):
        _ = list(map(int, _in[i+1].split(' ')))  # type:list(int)
        A_B_arr.append(_)
    # vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
    uf = Union_Find(N)
    for a, b in A_B_arr:
        a -= 1
        b -= 1
        uf.merge(a, b)
    cnt = Counter(uf.par)
    ans = cnt.most_common()[0][1]
    # ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    print(ans)


if __name__ == "__main__":
    main()