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 (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.annotation.tailrec

object Main extends App {


  val tmp = io.StdIn.readLine().split(' ').map(_.toInt)
  val (n, m) = (tmp(0), tmp(1))

  val tree = new IUnionFind(n)

  val t = io.StdIn.readLine().split(' ').map(_.toInt - 1)

  for(_ <- (1 to m)) {
    val ar = io.StdIn.readLine().split(' ').map(_.toInt - 1)
    val (x, y) = (ar(0), ar(1))
    tree.union(x, y)
  }

  val ans = t.zipWithIndex
    .count{
      case (x, i) => tree.connected(x, i)
    }

  println(ans)

  class IUnionFind(val size: Int) {

    private case class Node(var parent: Option[Int], var treeSize: Int)

    private val nodes = Array.fill[Node](size)(new Node(None, 1))

    def union(t1: Int, t2: Int): IUnionFind = {
      if (t1 == t2) return this

      val root1 = root(t1)
      val root2 = root(t2)
      if (root1 == root2) return this

      val node1 = nodes(root1)
      val node2 = nodes(root2)

      if (node1.treeSize < node2.treeSize) {
        node1.parent = Some(t2)
        node2.treeSize += node1.treeSize
      } else {
        node2.parent = Some(t1)
        node1.treeSize += node2.treeSize
      }
      this
    }

    def connected(t1: Int, t2: Int): Boolean = t1 == t2 || root(t1) == root(t2)

    @tailrec
    private def root(t: Int): Int = nodes(t).parent match {
      case None => t
      case Some(p) => root(p)
    }
  }
}