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 UnionFind0:
    def __init__(self, size):
        self.table = list(range(1,size+1))

    def find(self, x):
        return self.table[x-1]

    def union(self, x, y):
        x1 = self.find(x)
        y1 = self.find(y)
        if x1 == y1: return False
        # 併合
        for i in range(len(self.table)):
            if self.table[i] == y1: self.table[i] = x1
        return True

N,M = map(int,input().split())
P = list(map(int,input().split()))

DF = UnionFind0(N)

for m in range(M):
    a,b = map(int,input().split())
    DF.union(a,b) #O(N)
    
ans = 0
for i in range(1,N+1):
    #現在位置のグループ＝＝その位置にいる数字くんの行きたい位置のグループ
    if DF.find(i)==DF.find(P[i-1]): ans+=1
        
print(ans)