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: Runtime Error)

class UnionFind {
  constructor(n) {
    this.p = new Array(n).fill(null).map((_, i) => i)
    this.height = new Array(n).fill(0)
    this.size = new Array(n).fill(1)
  }

  unite(a, b) {
    this.link(this.findSet(a), this.findSet(b))
  }

  link(a, b) {
    if (a === b) return null

    if (this.height[a] > this.height[b]) {
      this.p[b] = a
      this.size[a] += this.size[b]
    } else {
      this.p[a] = b
      this.size[b] += this.size[a]

      if (this.height[a] === this.height[b]) {
        this.height[b]++
      }
    }
  }

  findSet(x) {
    if (this.p[x] === x) return x
    return this.findSet(this.p[x])
  }
}

const main = (input) => {
  input = input.trim().split('\n')
  const [n,] = input.shift().split(' ').map(Number)
  input = Array.from(new Set(input))

  const uf = new UnionFind(n+1)

  for (let i=0; i<input.length; i++) {
    const [a, b] = input[i].split(' ').map(Number)
    uf.unite(a, b)
  }

  // const ans = Math.max(...uf.size)
  let max = 0

  for (let i=0; i<uf.size.length; i++) max = Math.max(max, uf.size[i])
  return console.log(max)