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: Time Limit Exceeded)

def dfs(v, visited, graph)
  visited[v] = true
  target = graph[v].size.times.select{ |t| graph[v][t] == true }
    if target != []
      target.each{|x| dfs(x, visited, graph) if visited[x] == false }
    end
end

def main()
  n,m = gets.chomp.split(' ').map(&:to_i)
  target = gets.chomp.split(' ').map(&:to_i)
  dic = {}
  for i in 0..n-1 do
    dic.store(target[i],i)
  end

  graph = Array.new(n).map{Array.new(n,false)}
  a = Array.new(n,nil)
  b = Array.new(n,nil)
  ans = 0

  for num in 0..m-1 do
    x,y = gets.chomp.split(' ').map(&:to_i)
    a[num],b[num] = x-1, y-1
    graph[a[num]][b[num]] = true
    graph[b[num]][a[num]] = true
  end

  #solve
  for i in 1..n do
    visited = Array.new(n,false)
    dfs(dic[i], visited, graph)
    ans = ans + 1 if visited[i-1] == true
  end
  puts ans
end

main()