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)

#!/usr/bin/env python3
import sys

# refered to
# rank : http://at274.hatenablog.com/entry/2018/02/02/173000
# size : abc120 kaisetsu
class UnionFind:
    def __init__(self, n):
        # par = Parent Number or NoV
        self.par = [-1] * (n+1)
        # rank = Tree Height
        self.rank = [0] * (n+1)
    
    # 自分が所属する集合の数を返す
    def size(self, x):
        return -1*self.par[self.find(x)]
    
    def size_list(self):
        return [x*(-1) for x in self.par[1:] if x < 0]

    # 根なら番号を返す
    def find(self, x):
        
        if(self.par[x] < 1):
            return x
        self.par[x] = self.find(self.par[x])
        return self.par[x]

    # 親が同じか判定
    def same_check(self, x, y):
        return self.find(x) == self.find(y)

    def union(self, x, y):
        # 根を探す
        x = self.find(x)
        y = self.find(y)

        # もう繋がれている
        if(x == y):
            return False

        # つなぎ替える
        if(self.size(x) < self.size(y)):
            x,y = y,x
        
        # xのサイズを更新
        self.par[x] += self.par[y]
        # yのサイズ(おやばんごう)をxに
        self.par[y] = x

        return True 

def solve(N: int, M: int, p: "List[int]", x: "List[int]", y: "List[int]"):
    uf = UnionFind(N)

    for i in range(M):
        uf.union(x[i], y[i])

    ans = 0
    for i in range(1, N+1):
        if(uf.same_check(i, p[i-1])):
            ans += 1

    print(ans)

    return


# Generated by 1.1.3 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    p = [ int(next(tokens)) for _ in range(N) ]  # type: "List[int]"
    x = [int()] * (M)  # type: "List[int]" 
    y = [int()] * (M)  # type: "List[int]" 
    for i in range(M):
        x[i] = int(next(tokens))
        y[i] = int(next(tokens))
    solve(N, M, p, x, y)

if __name__ == '__main__':
    main()