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)

# https://www.leighhalliday.com/weighted-quick-union-find-algorithm-in-ruby
class UnionFind

  attr_accessor :nodes, :sizes

  def initialize(num)
    self.nodes = []
    self.sizes = []

    num.times do |n|
      self.nodes[n] = n
      self.sizes[n] = 1
    end
  end

  def root(i)
    # Loop up the chain until reaching root
    while nodes[i] != i do
      # path compression for future lookups
      nodes[i] = nodes[nodes[i]]
      i = nodes[i]
    end
    i
  end

  def union(i, j)
    rooti = root i
    rootj = root j

    # already connected
    return if rooti == rootj

    # root smaller to root of larger
    if sizes[i] < sizes[j]
      nodes[rooti] = rootj
      sizes[rootj] += sizes[rooti]
    else
      nodes[rootj] = rooti
      sizes[rooti] += sizes[rootj]
    end
  end

  def connected?(i, j)
    root(i) == root(j)
  end

end

N, M = gets.split.map &:to_i
P = gets.split.map &:to_i
xy = $<.map{|s|
  s.split.map &:to_i
}


uf = UnionFind.new(N)
xy.each{|x, y|
  uf.union(x-1, y-1)
}

p N.times.count{|i|
  uf.connected?(i,P[i]-1)
}