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: Time Limit Exceeded)

class UnionFind():
   def __init__(self, n):
       self.n = n
       self.parents = [-1] * n

   def find(self, x):
       if self.parents[x] < 0:
           return x
       else:
           self.parents[x] = self.find(self.parents[x])
           return self.parents[x]

   def union(self, x, y):
       x = self.find(x)
       y = self.find(y)

       if x == y:
           return

       if self.parents[x] > self.parents[y]:
           x, y = y, x

       self.parents[x] += self.parents[y]
       self.parents[y] = x

   def size(self, x):
       return -self.parents[self.find(x)]

   def same(self, x, y):
       return self.find(x) == self.find(y)

   def members(self, x):
       root = self.find(x)
       return [i for i in range(self.n) if self.find(i) == root]

   def roots(self):
       return [i for i, x in enumerate(self.parents) if x < 0]

   def group_count(self):
       return len(self.roots())

   def all_group_members(self):
       return {r: self.members(r) for r in self.roots()}

   def __str__(self):
       return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())


n,m = map(int,input().split())
pl = list(map(int,input().split()))
uf = UnionFind(n)
for _ in range(m):
   x,y = map(int,input().split())
   uf.union(x-1,y-1)
ans = 0
for member in uf.all_group_members().values():
   a = [pl[i]-1 for i in member]
   ans += len(set(a) & set(member))

print(ans)