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)

class UnionFind:
    def __init__(self, size):
        self.parent = [-1 for i in range(size)]#非負なら親ノード，負ならグループの要素数

    def root(self, x):
        if self.parent[x] < 0:
            return x
        else:
            self.parent[x] = self.root(self.parent[x]) #xをxの根に直接つなぐ
            return self.parent[x]

    def union(self, x, y):
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return False

        if self.parent[x] > self.parent[y]: #xの要素数がyの要素数より「小さい」とき
            x,y=y,x
        self.parent[x] += self.parent[y] #xの要素数を更新
        self.parent[y] = x #yをxにつなぐ
        return True

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



n,m = list(map(int,input().split()))
p = list(map(lambda x: int(x)-1,input().split()))

uf = UnionFind(n)
for i in range(m):
    x,y=list(map(int,input().split()))
    uf.union(x-1,y-1)

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

print(ans)