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: Wrong Answer)

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] p = new int[n];
		for (int i = 0; i < n; i++) {
			p[i] = sc.nextInt() - 1;
		}
		Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
		for (int i = 0; i < m; i++) {
			int x = sc.nextInt() - 1;
			int y = sc.nextInt() - 1;
			if (map.containsKey(x)) {
				if (map.containsKey(y)) {
					if (x != y) {
						Set<Integer> setX = map.get(x);
						Set<Integer> setY = map.get(y);
						if (setX.size() > setY.size()) {
							setX.addAll(setY);
							for (Integer o : setY) {
								map.put(o, setX);
							}
						} else {
							setY.addAll(setX);
							for (Integer o : setX) {
								map.put(o, setY);
							}
						}
					}
				} else {
					map.get(x).add(y);
					map.put(y, map.get(x));
				}
			} else {
				if (map.containsKey(y)) {
					map.get(y).add(x);
					map.put(x, map.get(y));
				} else {
					Set<Integer> set = new HashSet<Integer>();
					set.add(x);
					set.add(y);
					map.put(x, set);
					map.put(y, set);
				}
			}
		}
		sc.close();

		int ans = 0;
		for (int i = 0; i < n; i++) {
			if (p[i] == i) {
				ans++;
			} else if (map.get(i) == map.get(p[i])) {
				ans++;
			}
		}
		System.out.println(ans);
	}
}