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)

let inputString = "";
let currentLine = 0;
process.stdin.on("data", (data) => {
  inputString += data;
});
process.stdin.on("end", function () {
  inputString = inputString
    .trim()
    .split("\n")
    .map((str) => str.trim());
  main();
});
function readLine() {
  return inputString[currentLine++];
}
function main() {
  let [totalPeople, t] = readLine()
    .split(" ")
    .map((n) => parseInt(n));
  const graph = {};
  const seenCouple = {};
  while (t--) {
    let [p1, p2] = readLine()
      .split(" ")
      .map((n) => parseInt(n));

    if (!seenCouple[`${p1}-${p2}`]) {
      seenCouple[`${p1}-${p2}`] = true;

      if (!graph[p1]) graph[p1] = { [p2]: p2 };
      else graph[p1] = { ...graph[p1], ...{ [p2]: p2 } };
      if (!graph[p2]) graph[p2] = { [p1]: p1 };
      else graph[p2] = { ...graph[p2], ...{ [p1]: p1 } };
    }
  }

  let max = 0;
  let visited = {};
  for (let key in graph) {
    max = Math.max(max, dfs(key));
  }
  console.log(max);

  function dfs(node) {
    let count = 0;
    if (!visited[node]) {
      visited[node] = true;
      count++;
      const neighbours = graph[node];
      for (let neighbour in neighbours) {
        if (!visited[neighbour]) {
          count += dfs(neighbour);
        }
      }
    }
    return count;
  }
}