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)

n, M = map(int, input().split())
AB = [list(map(int, input().split())) for _ in range(M)]


def bellman_ford(s):
    d = [float('inf')]*n # 各頂点への最小コスト
    d[s] = 0 # 自身への距離は0
    for i in range(n):
        update = False # 更新が行われたか
        for x, y, z in g:
            if d[y] > d[x] + z:
                d[y] = d[x] + z
                update = True
        if not update:
            break
        # 負閉路が存在
        if i == n - 1:
            return -1
    return d


# n, w = [int(x) for x in input().split()] # n:頂点数, w:辺の数
g = []
for a, b in AB:
    # x, y, z = [int(x) for x in input().split()] # 始点,終点,コスト
    g.append([a-1, b-1, 1])
    g.append([b-1, a-1, 1]) # 有向グラフでは削除
res = bellman_ford(0)
print(len([n for r in res if r != float('inf')]))