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
input=sys.stdin.readline

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

    def find(self,x): #根を見つける、繋ぎ直す
        if self.parents[x]<0:
            return x
        else:
            self.parents[x]=self.find(self.parents[x])
            return self.parents[x]
    
    def unite(self,x,y): #x,yの含むグループを併合する
        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 same(self,x,y):#xとyが同じグループにいるか判定
        return self.find(x)==self.find(y)
    
    def members(self,x):#xと同じグループのメンバーを列挙
        root=self.find(x)
        return [i for i in range(self.n) if self.find(i)==root]
    
    def size(self,x):#xが属するグループのメンバーの数
        return -self.parents[self.find(x)]
    
    def roots(self):#ufの根を列挙
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):#グループの数を数える
        return len(self.roots())

    def all_group_members(self):#根:メンバの辞書を返す
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):#print()での表示用
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def main():
    n,m = map(int,input().split())
    p = [int(i)-1 for i in input().split()]
    dic = {i:idx for idx,i in enumerate(p)}   

    uf = UnionFind(n)
    for _ in range(m):
        x,y = map(int,input().split())
        x -= 1
        y -= 1
        uf.unite(x,y)
    
    cnt = 0
    for i in range(n):
        if uf.same(dic[i],i):
            cnt += 1
    
    print(cnt)  
  
if __name__=='__main__':
    main()