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 UnionFindWithRank
  def initialize(n)
    @parent = Array.new(n, -1)
    @rank = Array.new(n, 0)
  end

  def root(x)
    return x if @parent[x] == -1
    @parent[x] = root(@parent[x])
    return @parent[x]
  end

  def same?(x, y)
    return root(x) == root(y)
  end

  def merge(x, y)
    x, y = root(x), root(y)
    return false if x == y
    x, y = y, x if @rank[x] < @rank[y]
    @rank[x] += 1 if @rank[x] == @rank[y]
    @parent[y] = x
    return true
  end
end

N, M = gets.split.map(&:to_i)
p = gets.split.map(&:to_i)
uf = UnionFindWithRank.new(N)
M.times do
  x,y = gets.split.map(&:to_i).map{|e| e-1}
  uf.merge(p[x]-1,p[y]-1)
end
ans = 0
N.times do |i|
  ans += 1 if uf.same?(i, p[i]-1)
end
puts ans