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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

const MOD int64 = int64(1e9+7)
var sc = bufio.NewScanner(os.Stdin)

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

var tonari = make([][]int, 200010)

func dfs( node int, connected []bool) int {
	ret := 0
	for _, i := range tonari[node] {
		if connected[i]	{
			continue
		}
		connected[i] = true
		ret += dfs(i, connected)+1
	}
	return ret
}

func main() {
	sc.Split(bufio.ScanWords)
	N := nextInt()
	M := nextInt()
	connected := make([]bool, N, N)
	var a,b int
	for i := 0; i < M; i++ {
		a = nextInt()
		b = nextInt()
		a--; b--
		tonari[a] = append(tonari[a],b)
		tonari[b] = append(tonari[b],a)
	}
	ans := 1
	for i:=0; i<N; i++ {
		if connected[i] {
			continue
		}
		if cnt := dfs(i, connected); ans < cnt{
			ans = cnt
		}
	}
	fmt.Printf("%d\n", ans)
}