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: Wrong Answer)

gets
ps = [0] + gets.split.map(&:to_i)
xys = $stdin.read.split.each_slice(2).to_a.map {|a| a.map(&:to_i)}


a = []
h = {}
idx = 0

xys.each do |x,y|
  if !h[x] && !h[y]
    h[x] = h[y] = idx
    a << [x,y]
    idx += 1
  elsif h[x] && h[y] && a[h[x]].equal?(a[h[y]])
    a[h[x]] << x
    a[h[x]] << y
  elsif h[x] && h[y]
    # merge
    a_ = []
    a_ += a[h[x]]
    a_ += a[h[y]]
    a[h[x]] = a_
    a[h[y]] = a_
  elsif h[x] && !h[y]
    h[y] = h[x]
    a[h[x]] << y
  elsif !h[x] && h[y]
    h[x] = h[y]
    a[h[y]] << x
  end
end

p a.uniq
p h

ret = 0

(1..ps.length).each do |i|
  ret += 1 if !h[i] && ps[i] == i
end

a.uniq.each do |t|
  ret += (t.map {|i| ps[i]} & t).count
end

p ret