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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

const input = require('fs').readFileSync('/dev/stdin', 'utf8')
const lines = input.split('\n')
let [N, M] = lines[0].trim().split(' ').map(e => parseInt(e))

const g = {}
let ai = 0
const map = {}
for (let i = 1; i <= M; i++) {
    const [a, b] = lines[i].trim().split(' ').map(e => parseInt(e))
    if (!map[a] && !map[b]) {
        ai++
        g[ai.toString()] = [a, b]
        map[a] = ai.toString()
        map[b] = ai.toString()
    } else if (map[a] && map[b]) {
        if (map[a] !== map[b]) {
            if (g[map[a]].length > g[map[b]].length) {
                const _b = map[b]
                g[map[b]].forEach(n => {
                    g[map[a]].push(n)
                    map[n] = map[a]
                })
                delete g[_b]
            } else {
                const _a = map[a]
                g[map[a]].forEach(n => {
                    g[map[b]].push(n)
                    map[n] = map[b]
                })
                delete g[_a]
            }
        }
    } else if (map[a]) {
        g[map[a]].push(b)
        map[b] = map[a]
    } else if (map[b]) {
        g[map[b]].push(a)
        map[a] = map[b]
    }
}

let ans = 0
Object.keys(g).forEach(key => {
    ans = Math.max(ans, g[key].length)
})
console.log(ans)