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: N, M = list(map(int, input().split()))
  p = list(map(int, input().split()))
  x = []
  y = []
  for _ in range(M):
      xi, yi = list(map(int, input().split()))
      x.append(xi)
      y.append(yi)
  
  def count(l):
      out = 0
      for i, v in enumerate(l):
          if i + 1 == v:
              out += 1
      return out
  
  def swap(l, a, b):
      va = l[a]
      vb = l[b]
      l1 = l.copy()
      l1[a] = vb
      l1[b] = va
      return l1
  
  depth = 0
  max_count = count(p)
  max_d = M**(N*np.log(N))
  children =	[p]
  next_children =	[]
  while depth <= max_d:
      for child in children:
          for i in range(M):
  	    pi = swap(child)
              ci = count(pi)
              if ci > max_count:
  		max_count = ci
  	    next_children.append(pi)
      depth += 1
      children = next_children
      next_chldren = []
      if max_count = len(p):
  	break
  
  print(max_count)
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."}]