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 PyPy3 (7.3.0) (judge verdict: Accepted)

class UnionFind:
    def __init__(self, num):
        self.parent = [i for i in range(num + 1)]

    def find(self, node):
        if self.parent[node] == node:
            return node

        self.parent[node] = self.find(self.parent[node])
        return self.parent[node]

    def union(self, node1, node2):
        node1 = self.find(node1)
        node2 = self.find(node2)

        if node1 == node2:
            return

        if self.parent[node1] > self.parent[node2]:
            node1, node2 = node2, node1

        self.parent[node2] = node1
        return

    def same(self, node1, node2):
        return self.find(node1) == self.find(node2)


n, m = map(int, input().split())
p = list(map(int, input().split()))

uf = UnionFind(n)
for _ in range(m):
    x, y = map(int, input().split())
    x -= 1
    y -= 1
    uf.union(p[x], p[y])

ans = 0
for i in range(n):
    if p[i] == i + 1 or uf.same(p[i], i + 1):
        ans += 1

print(ans)