output_description: Print the maximum possible number of i such that p_i = i after operations.
  
  * * *
input_description: 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
attempt: ###############################################################################
  
  from sys import stdout
  from bisect import bisect_left as binl
  from copy import copy, deepcopy
  from collections import defaultdict
  
  
  mod = 1
  
  
  def intin():
      input_tuple = input().split()
      if len(input_tuple) <= 1:
          return int(input_tuple[0])
      return tuple(map(int, input_tuple))
  
  
  def intina():
      return [int(i) for i in input().split()]
  
  
  def intinl(count):
      return [intin() for _ in range(count)]
  
  
  def modadd(x, y):
      global mod
      return (x + y) % mod
  
  
  def modmlt(x, y):
      global mod
      return (x * y) % mod
  
  
  def lcm(x, y):
      while y != 0:
          z = x % y
          x = y
          y = z
      return x
  
  
  def combination(x, y):
      assert x >= y
  
      if y > x // 2:
          y = x - y
  
      ret = 1
  
      for i in range(0, y):
          j = x - i
          i = i + 1
  
          ret = ret * j
          ret = ret // i
  
      return ret
  
  
  def get_divisors(x):
      retlist = []
      for i in range(1, int(x**0.5) + 3):
          if x % i == 0:
              retlist.append(i)
              retlist.append(x // i)
      return retlist
  
  
  def get_factors(x):
      retlist = []
      for i in range(2, int(x**0.5) + 3):
          while x % i == 0:
              retlist.append(i)
              x = x // i
      retlist.append(x)
      return retlist
  
  
  def make_linklist(xylist):
      linklist = {}
      for a, b in xylist:
          linklist.setdefault(a, [])
          linklist.setdefault(b, [])
          linklist[a].append(b)
          linklist[b].append(a)
      return linklist
  
  
  def calc_longest_distance(linklist, v=1):
      distance_list = {}
      distance_count = 0
      distance = 0
      vlist_previous = []
      vlist = [v]
      nodecount = len(linklist)
  
      while distance_count < nodecount:
          vlist_next = []
          for v in vlist:
              distance_list[v] = distance
              distance_count += 1
              vlist_next.extend(linklist[v])
          distance += 1
          vlist_to_del = vlist_previous
          vlist_previous = vlist
          vlist = list(set(vlist_next) - set(vlist_to_del))
  
      max_distance = -1
      max_v = None
      for v, distance in distance_list.items():
          if distance > max_distance:
              max_distance = distance
              max_v = v
  
      return (max_distance, max_v)
  
  
  def calc_tree_diameter(linklist, v=1):
      _, u = calc_longest_distance(linklist, v)
      distance, _ = calc_longest_distance(linklist, u)
      return distance
  
  
  class UnionFind:
  
      def __init__(self, n):
          self.parent = [i for i in range(n)]
  
      def root(self, i):
          if self.parent[i] == i:
              return i
          self.parent[i] = self.root(self.parent[i])
          return self.parent[i]
  
      def unite(self, i, j):
          rooti = self.root(i)
          rootj = self.root(j)
          if rooti == rootj:
              return
          if i < j:
              self.parent[rootj] = rooti
          else:
              self.parent[rooti] = rootj
  
      def same(self, i, j):
          return self.root(i) == self.root(j)
  
  
  ###############################################################################
  
  
  def main():
      n, m = intin()
      plist = intina()
      xylist = intinl(m)
  
      new_xylist = []
      for x, y in xylist:
          if x < y:
              new_xylist.append((x, y))
              continue
          new_xylist.append((y, x))
      new_xylist.sort()
      xylist = new_xylist
  
      u = UnionFind(n)
  
      for x, y in xylist:
          u.unite(x - 1, y - 1)
  
      ans = 0
  
      for i, p in enumerate(plist):
          if u.same(i, p - 1):
              ans += 1
  
      print(ans)
  
  
  if __name__ == "__main__":
      main()
  
problem_description: 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.
samples: [{"input": "5 2\n    5 3 1 4 2\n    1 3\n    5 4", "output": "2\n    \n\nIf we perform the operation by choosing j=1, p becomes `1 3 5 4 2`, which is\noptimal, so the answer is 2.\n\n* * *"}, {"input": "3 2\n    3 2 1\n    1 2\n    2 3", "output": "3\n    \n\nIf we perform the operation by, for example, choosing j=1, j=2, j=1 in this\norder, p becomes `1 2 3`, which is obviously optimal. Note that we may choose\nthe same j any number of times.\n\n* * *"}, {"input": "10 8\n    5 3 6 8 7 10 9 1 2 4\n    3 1\n    4 1\n    5 9\n    2 5\n    6 5\n    3 5\n    8 9\n    7 9", "output": "8\n    \n\n* * *"}, {"input": "5 1\n    1 2 3 4 5\n    1 5", "output": "5\n    \n\nWe do not have to perform the operation."}]