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)

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# FileName: 	D
# CreatedDate:  2020-08-29 21:38:12 +0900
# LastModified: 2020-08-29 21:54:27 +0900
#


import os
import sys
sys.setrecursionlimit(900000)
# import numpy as np
# import pandas as pd
from collections import Counter


def dfs(u, color, path, n, d):
    color[u] = d
    for v in path[u]:
        if color[v] == -1:
            dfs(v, color, path, n, d)



def main():
    n, m = map(int, input().split())
    path = [[] for _ in range(n+1)]
    for _ in range(m):
        a, b = map(int, input().split())
        path[a].append(b)
        path[b].append(a)
    color = [-1]*(n+1)
    d = 0
    for i in range(1, n+1):
        if color[i] == -1:
            dfs(i, color, path, n, d)
            d += 1
    color = color[1:]
    color_cnt = Counter(color).values()
    print(max(color_cnt))



if __name__ == "__main__":
    main()