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: Accepted)

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

const A = []
const B = []
for (let i = 0; i < M; i++) {
    const l = lines[i + 1].split(' ').map(Number)
    A[i] = l[0]
    B[i] = l[1]
}

const g = {}
for (let i = 0; i < M; i++) {
    const l = [A[i], B[i]]
    if (!g[A[i]] && !g[B[i]]) {
        g[A[i]] = l
        g[B[i]] = l
    } else if (g[A[i]] && g[B[i]]) {
        if (g[A[i]] !== g[B[i]]) {
            if (g[A[i]].length < g[B[i]].length) {
                g[A[i]].forEach(n => {
                    g[B[i]].push(n)
                    g[n] = g[B[i]]
                })
            } else {
                g[B[i]].forEach(n => {
                    g[A[i]].push(n)
                    g[n] = g[A[i]]
                })
            }
        }
    } else if (g[A[i]]) {
        if (!g[B[i]]) {
            g[B[i]] = g[A[i]]
        }
        g[B[i]].push(B[i])
    } else if (g[B[i]]) {
        if (!g[A[i]]) {
            g[A[i]] = g[B[i]]
        }
        g[A[i]].push(A[i])
    }
}

let ans = 1
for (let key in g) {
    ans = Math.max(ans, g[key].length)
}
console.log(ans)