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 Haskell (GHC 8.8.3) (judge verdict: Runtime Error)

import           Control.Monad
import qualified Data.ByteString as B
import           Data.List
import qualified Data.Set        as S
import qualified Data.Vector     as V

readInt :: IO [Int]
readInt = map read . words <$> getLine

readIntlines :: Int -> IO [[Int]]
readIntlines n = map (map read . words) <$> replicateM n getLine

main :: IO ()
main = do
    [n, m] <- readInt
    ab <- readIntlines m
    print $ solve ab

solve :: [[Int]] -> Int
solve xs = maximum $ map S.size $ foldl' (\acc x -> set x acc) [] xs

set :: [Int] -> [S.Set Int] -> [S.Set Int]
set y gs = let (ts, fs) = partition (\c -> or $ map (`S.member` c) y) gs
            in if null ts
                  then (S.fromList y) : fs
                  else S.unions (S.fromList y : ts) : fs