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)

# -*- coding: utf-8 -*-
# モジュールのインポート
from collections import defaultdict, deque


def get_input() -> tuple:
    """
    標準入力を取得する.

    Returns:\n
        tuple: 標準入力
    """
    # 標準入力を取得
    N, M = list(map(int, input().split()))
    G = defaultdict(set)
    for m in range(M):
        A_i, B_i = list(map(int, input().split()))
        G[A_i - 1].add(B_i - 1)
        G[B_i - 1].add(A_i - 1)

    return N, M, G


def main(N: int, M: int, G: dict) -> None:
    """
    メイン処理.

    Args:\n
        N (int): 人数(2 <= N <= 2 * 10**5)
        M (int): 情報の数(0 <= M <= 2 * 10**5)
        G (dict): 情報
    """
    # 求解処理
    seen = [False for i in range(N)]

    ans = 0
    for i in range(N):
        if seen[i]:
            continue
        friends = 0
        que = deque()

        seen[i] = True
        que.append(i)
        friends += 1

        while len(que) > 0:
            v = que.popleft()
            for w in G[v]:
                if not seen[w]:
                    seen[w] = True
                    que.append(w)
                    friends += 1

        ans = max(ans, friends)

    # 結果出力
    print(ans)


if __name__ == "__main__":
    # 標準入力を取得
    N, M, G = get_input()

    # メイン処理
    main(N, M, G)