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.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main implements Runnable {

    private static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 256 * 1024 * 1024).start();
    }

    public void run() {
        int n = sc.nextInt(), m = sc.nextInt();
        int p[] = new int[n + 1];
        Map<Integer, List<Integer>> neighbors = new HashMap<>();
        for (int i = 1; i <= n; i++) {
            p[i] = sc.nextInt();
            neighbors.put(p[i], new ArrayList<>());
        }
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt(), b = sc.nextInt();
            neighbors.get(a).add(b);
            neighbors.get(b).add(a);
        }

        Map<Integer, Integer> labels = BFS.labelLinking(neighbors);
        int count = 0;
        for (int i = 1; i < n + 1; i++) {
            if (labels.containsKey(i) && labels.get(i).equals(labels.get(p[i]))) {
                count++;
            }
        }
        System.out.println(count);
    }
}

class BFS {

    static Map<Integer, Integer> labelLinking(Map<Integer, List<Integer>> graph) {
        Map<Integer, Integer> labels = new HashMap<>();
        int linkId = 1;
        for (Map.Entry<Integer, List<Integer>> vertex : graph.entrySet()) {
            if (!labels.containsKey(vertex.getKey())) {
                _labelLinking(graph, labels, linkId, vertex.getKey());
                linkId++;
            }
        }
        return labels;
    }

    private static void _labelLinking(Map<Integer, List<Integer>> graph,
                                      Map<Integer, Integer> labels, int linkId, int root) {
        Deque<Integer> queue = new ArrayDeque<>();
        queue.add(root);
        labels.put(root, linkId);
        while (!queue.isEmpty()) {
            int parent = queue.poll();
            for (Integer child : graph.get(parent)) {
                if (!labels.containsKey(child)) {
                    labels.put(child, linkId);
                    queue.add(child);
                }
            }
        }
    }
}