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
  # property parent : Int32
  # property s : Int32
  # property r : Int32
  attr_accessor :parent, :s, :r # Ruby
    
  def initialize(n)
    @parent = -1
    @s = 1
    @r = 0
  end
end

class UnionFindTree
  
  # property nodes : Array(Node) # Crystal
  attr_accessor :nodes        # Ruby

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

  def root(x)
    t = nodes[x].parent
    return x if t < 0
    nodes[x].parent = root(t)
  end
  
  def unite(a, b)
    a = root(a)
    b = root(b)
    return false if a == b
    a, b = b, a if nodes[a].r < nodes[b].r
    nodes[a].r += 1 if nodes[a].r == nodes[b].r # by rank
    nodes[a].s += nodes[b].s                    # by size
    nodes[a].parent += nodes[b].parent # if negative value means size
    nodes[b].parent = a
    true
  end
  
  def unite_by_size(a, b)
    a = root(a)
    b = root(b)
    return false if a == b
    a, b = b, a if nodes[a].s < nodes[b].s
    nodes[b].parent = a
    nodes[a].s += nodes[b].s
    true
  end
  
  def unite_by_rank(a, b)
    a = root(a)
    b = root(b)
    return false if a == b
    a, b = b, a if nodes[a].r < nodes[b].r
    nodes[b].parent = a
    nodes[a].r += 1 if nodes[a].r == nodes[b].r
  end

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

  def size(node)
    -nodes[root(node)]
  end
  
  def parents; nodes.map{|i|i.parent} end
  def r; nodes.map{|i|i.r} end
  def s; nodes.map{|i|i.s} end
  #def diff_weight; @nodes.map{|i|i.diff_weight) end
end

n, m = gets.to_s.chomp.split.map{|i| i.to_i}
perm = gets.to_s.chomp.split.map{|i| i.to_i - 1 }
xys  = Array.new(m){ gets.to_s.chomp.split.map{|i| i.to_i - 1 } }

uf = UnionFindTree.new(n)
xys.each{|x,y| uf.unite(x,y); }
ans = (0...n).count{ |i| uf.same?(i, perm[i]) }

puts ans