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():
    # 作りたい要素数nで初期化
    # 使用するインスタンス変数の初期化
    def __init__(self, n):
        self.n = n
        # root[x]<0ならそのノードが根かつその値が木の要素数
        # rootノードでその木の要素数を記録する
        self.root = [-1]*(n+1)
        # 木をくっつける時にアンバランスにならないように調整する
        self.rnk = [0]*(n+1)

    # ノードxのrootノードを見つける
    def Find_Root(self, x):
        if(self.root[x] < 0):
            return x
        else:
            # ここで代入しておくことで、後の繰り返しを避ける
            self.root[x] = self.Find_Root(self.root[x])
            return self.root[x]
    # 木の併合、入力は併合したい各ノード

    def Unite(self, x, y):
        # 入力ノードのrootノードを見つける
        x = self.Find_Root(x)
        y = self.Find_Root(y)
        # すでに同じ木に属していた場合
        if(x == y):
            return
        # 違う木に属していた場合rnkを見てくっつける方を決める
        elif(self.rnk[x] > self.rnk[y]):
            self.root[x] += self.root[y]
            self.root[y] = x

        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            # rnkが同じ（深さに差がない場合）は1増やす
            if(self.rnk[x] == self.rnk[y]):
                self.rnk[y] += 1
    # xとyが同じグループに属するか判断

    def isSameGroup(self, x, y):
        return self.Find_Root(x) == self.Find_Root(y)

    # ノードxが属する木のサイズを返す
    def Count(self, x):
        return -self.root[self.Find_Root(x)]


N, M = map(int, input().split())
p = [int(i) for i in input().split()]
uf = UnionFind(N)

for _ in range(M):
    x, y = map(int, input().split())
    uf.Unite(x, y)

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