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)

function MainD(input){
    const arr = input.split('\n');
    const arrAB = arr.map(val => val.split(' '));
    const setAB = arr.map(val => new Set(val.split(' ')));
    // グループ数
    let groupNum = arrAB.length;
    for(let i=1; i<arrAB.length; i++){ //まず2つめから
        for(let j=0; j<i; j++){ // 1つ目をみる
            if(setAB[j].has(arrAB[i][0]) || setAB[j][0].has(arrAB[i][1])){
                groupNum--;
                break;
            }
        }
    }
    console.log(groupNum);
}
MainD(require("fs").readFileSync("/dev/stdin", "utf8").trim());