Problem:
Score : 400 points

Problem Statement

We have a permutation of the integers from 1 through N, p_1, p_2, .., p_N.
We also have M pairs of two integers between 1 and N (inclusive), represented as (x_1,y_1), (x_2,y_2), .., (x_M,y_M).
AtCoDeer the deer is going to perform the following operation on p as many times as desired so that the number of i (1 ≤ i ≤ N) such that p_i = i is maximized:

Choose j such that 1 ≤ j ≤ M, and swap p_{x_j} and p_{y_j}.

Find the maximum possible number of i such that p_i = i after operations.

Constraints

2 ≤ N ≤ 10^5

1 ≤ M ≤ 10^5

p is a permutation of integers from 1 through N.

1 ≤ x_j,y_j ≤ N

x_j ≠ y_j

If i ≠ j, \{x_i,y_i\} ≠ \{x_j,y_j\}.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
p_1 p_2 .. p_N
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the maximum possible number of i such that p_i = i after operations.

Sample Input 1

5 2
5 3 1 4 2
1 3
5 4

Sample Output 1

2

If we perform the operation by choosing j=1, p becomes 1 3 5 4 2, which is optimal, so the answer is 2.

Sample Input 2

3 2
3 2 1
1 2
2 3

Sample Output 2

3

If we perform the operation by, for example, choosing j=1, j=2, j=1 in this order, p becomes 1 2 3, which is obviously optimal.
Note that we may choose the same j any number of times.

Sample Input 3

10 8
5 3 6 8 7 10 9 1 2 4
3 1
4 1
5 9
2 5
6 5
3 5
8 9
7 9

Sample Output 3

8

Sample Input 4

5 1
1 2 3 4 5
1 5

Sample Output 4

5

We do not have to perform the operation.

Solution in Scala (2.13.1) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

object Main extends App {
  val sc = new Scanner(System.in);

  val n, m = sc.nextInt();

  val ps = List.fill(n)(sc.nextInt() - 1);

  val xy = List.fill(m)((sc.nextInt() - 1, sc.nextInt() - 1));

  val root = (0 until n).toArray;

  def fetchRoot(i: Int): Int = {
    val r = root(i);
    if (r == i) {
      r;
    } else {
      val r2 = fetchRoot(r);
      root(i) = r2;
      r2;
    }
  }

  xy.foreach { case (x, y) =>
    val rx = fetchRoot(x);
    val ry = fetchRoot(y);
    if (rx != ry) {
      val (rx2, ry2) = if (rx < ry) {
        (rx, ry);
      } else {
        (ry, rx);
      }
      root(ry2) = rx2;
    }
  }

  val answer = (0 until n).filter { i =>
    val p = ps(i);
    val rp = fetchRoot(p);
    val ri = fetchRoot(i);
    ri == rp;
  }.size;

  println(answer);
}