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 Ruby (2.3.3) (judge verdict: Accepted)

class Node
  attr_accessor :parent, :rank

  def initialize(n)
    @parent = n
    @rank = 0
  end
end

class UnionFindTree
  def initialize(n)
    @nodes = (0..n).to_a.map { |i| Node.new(i) }
  end

  def find(x)
    return x if @nodes[x].parent == x

    @nodes[x].parent = find(@nodes[x].parent)
  end

  def unite(a, b)
    a = find(a)
    b = find(b)
    return if a == b

    if @nodes[a].rank < @nodes[b].rank
      @nodes[a].parent = b
    else
      @nodes[b].parent = a
      @nodes[a].rank += 1 if @nodes[a].rank == @nodes[b].rank
    end
  end

  def same?(a, b)
    find(a) == find(b)
  end
end

def main
  n, m = ARGF.gets.split.map(&:to_i)
  # puts("Error") if n <= 0

  seq = ARGF.gets.split.map(&:to_i)
  # pp seq

  # グループ分け
  nft = UnionFindTree.new(n)
  m.times do
    p0, p1 = ARGF.gets.split.map(&:to_i)
    nft.unite(p0, p1)
  end

  match_num = 0
  pos = 1
  seq.each do |num|
    if num == pos
      # 既に所定の位置にある場合は即インクリメント
      match_num += 1

    elsif nft.same?(pos, num)
      # pickした数字を所定の位置に持っていけるか？
      # 現在のpos数 と pickした数字 が同じGroup なら持っていける
      match_num += 1
    end

    pos += 1
  end

  puts match_num
end

if __FILE__ == $0
  main
end