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 UnionFind
  def initialize(n); @data = Array.new(n,-1); end
  def find(a); @data[a] < 0 ? a : @data[a] = find(@data[a]); end
  def same?(a,b); find(a) == find(b); end
  def size(a); -@data[find(a)]; end
  def unite(a,b)
    a = find(a); b = find(b)
    return if a == b
    a,b = b,a if @data[a] > @data[b]
    @data[a] += @data[b]; @data[b] = a
  end
end

# UnionFind.new(N) := 要素数Nで初期化
# Ai := i番目の要素
# find(i) := Aiが属するグループ
# same?(i,j) := Ai,Ajが同じグループに属する
# size(i) := Aiが属するグループの要素の数
# unite(i,j) := Ai,Ajを同じグループに統合する

n,m = gets.split.map(&:to_i)
a = gets.split.map(&:to_i).map{|v|v-1}

uf = UnionFind.new(n)
m.times do
  x,y = gets.split.map(&:to_i)
  x -= 1; y -= 1
  uf.unite(x,y)
end

ans = 0
n.times do |i|
  ans += 1 if uf.same?(a[i],i)
end

puts ans