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 Python (3.4.3) (judge verdict: Accepted)

import sys

N, M = map(int, input().split())
p = list(map(int, input().split()))
p.insert(0,0)
pairs = [[int(c) for  c in l.strip().split()] for l in sys.stdin]

link = [[] for _ in  range(N+1)]
for i in range(M):
  x = pairs[i][0]
  y = pairs[i][1]
  link[x].append(y)
  link[y].append(x)


ans = 0
visited = set()
#数列の１つ１つの数字について
for i in range(1, N+1):

  #既に1度見た数字は飛ばす
  if i in visited:
    continue
  
  #linkedに交換可能なひとまとまりの数字をセットとして格納
  #ここでlinkedをリセット
  linked = set()

  #todoにはリンク先を調べないといけない数字のリストを格納
  todo = [i]
  while todo:

    #todoから1つずつ
    cur = todo.pop()
    for dest in link[cur]:

      #既にグループに入っている数字は飛ばす
      if dest not in linked:
        linked.add(dest)
        todo.append(dest)

  #「位置」を表すグループと「数字」を表すグループのオーバーラップ部分を数える
  ans += len(linked & set(p[n] for n in linked))
  #一度見た数字は再訪しない
  visited |= linked

#最後にまだ一度も見ていない数字のなかで、i = p[i]であるものを探す
for i in set(range(1,N+1)) - visited:
  ans += i == p[i]

print(ans)