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: Accepted)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N, M = in.next().toInt
  val p = new Array[Int](N+1)
  val swaps = new Array[(Int, Int)](M)

  for (i <- 1 to N) {
    p(i) = in.next().toInt
  }
  for(i <- 0 until M){
    val a, b = in.next().toInt
    swaps(i) = (a, b)
  }


  //
  val uf = new UnionFind(N+1)

  for ((a, b) <- swaps) {
    uf.connect(a, b)
  }

  var ans = 0
  for(i <- 1 to N) {
    if(uf.isConnect(p(i), i))
      ans += 1
  }

  println(ans)
}


class UnionFind(v: Int){
  // 根であれば *そのグループの要素数(負)* が、子であれば親の番号が入る。
  val uni = Array.fill(v)(-1)

  // 頂点 v の所属するグループを調べる
  def root(v: Int): Int =
    uni(v) < 0 match{
      case true  =>  // v が親の場合
        return v
      case false =>  // v が子の場合
        uni(v) = root(uni(v))  // 親のrootを調べる
        return uni(v)
    }

  // 頂点 a と頂点 b をつなぐ。もともと同じグループのとき、False を返す
  def connect(a: Int, b: Int): Boolean = {
    // まずはそれぞれ根の番号に置き換える
    var ra = root(a)
    var rb = root(b)
    if(ra == rb)
      return false  // a と b がそもそも同じグループに属しているなら即終了

    // ra を大きなグループにしたいので、逆であれば入れ替える
    if(uni(ra) > uni(rb)) {  // rbの方が要素数が多ければ
      var tmp = ra
      ra = rb
      rb = tmp
    }

    // ra と rb を結合し、rb の親を ra とする
    uni(ra) += uni(rb)
    uni(rb) = ra
    return true
  }

  // 頂点 a, b が同じグループであるかを調べる
  def isConnect(a: Int, b: Int): Boolean =
    root(a) == root(b)

  // 頂点 a を含むグループの頂点数を調べる
  def size(a: Int): Int = -uni(root(a))
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}