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 max_fixed_point(edges,perm)
  ret = 0
  n = edges.size
  visited = Array.new(n)
  no = 0
  n.times{|i|
    next if visited[i]
    queue = [i]
    idx_set=[]
    perm_set=[]
    until queue.empty?
      v = queue.shift
      visited[v] = no
      idx_set << v
      perm_set << perm[v]-1
      edges[v].each{|w|
        next if visited[w]
        queue.push(w)
      }
    end
    no += 1
    ret += (idx_set&perm_set).size
  }
  ret
end

n,m=gets.split.map(&:to_i)
perm=gets.split.map(&:to_i)

edges=Array.new(n){Array.new}
m.times{
  x,y=gets.split.map(&:to_i)
  x-=1;y-=1
  edges[x] << y
  edges[y] << x
}

 puts max_fixed_point(edges,perm)