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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
static class UnionFind {
    List<Integer> Parent;
    UnionFind(int N) {
        Parent = new ArrayList<Integer>();
        Integer[] values = new Integer[N];
        Arrays.fill(values, -1);
        Parent.addAll(Arrays.asList(values));
    }
    int root(int A) {
        if (Parent.get(A) < 0)
            return A;
        int root = root(Parent.get(A));
        Parent.set(A, root);
        return root;
    }
    int size(int A) {
        return -Parent.get(root(A));
    }

    boolean connect(int A, int B) {
        A = root(A);
        B = root(B);
        if (A == B) {
            return false;
        }

        if (size(A) < size(B)) {
            int temp = A;
            A = B;
            B = temp;
        }

        Parent.set(A, Parent.get(A) + Parent.get(B));
        Parent.set(B, A);

        return true;
    }
}
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int M = Integer.parseInt(sc.next());
        int[] p = new int[N];
        for (int i=0;i<N;i++) {
            p[i] = Integer.parseInt(sc.next());
        }
        UnionFind uf = new UnionFind(N);
        for (int i=0;i<M;i++) {
            int a = Integer.parseInt(sc.next())-1;
            int b = Integer.parseInt(sc.next())-1;
            uf.connect(a, b);
        }
        int ans = 0;
        for (int i=0;i<N;i++) {
            if (uf.root(p[i]-1)==uf.root(i)) {
                ans++;
            }
        }
        System.out.println(ans);
    }

    public static int kyotsu(ArrayList<Integer> a, ArrayList<Integer> b) {
        int ans = 0;
        int j=0;
        for (int i=0;i<a.size();i++) {
            if (a.get(i)==b.get(j)) {
                ans++;
                j++;
                if (j==b.size()) {
                    break;
                }
            } else if (a.get(i)<b.get(j)) {
                continue;
            } else if (a.get(i)>b.get(j)) {
                i--;
                j++;
                if (j==b.size()) {
                    break;
                }
            }
        }
        return ans;
    }
}