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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
	public static Map<Integer, ArrayList<Integer>> map;
	public static int N;
	public static void main(String[] args) {
		Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		N = scan.nextInt();
		int M = scan.nextInt();
		Integer []p = new Integer[N];
		int []x = new int[M];
		int []y = new int[M];
		for(int i = 0; i < N; i++) {
			p[i] = scan.nextInt();
		}
		for(int i = 0; i < M; i++) {
			x[i] = scan.nextInt();
			y[i] = scan.nextInt();
		}
		scan.close();

		// 隣接リスト
		map = new TreeMap<Integer, ArrayList<Integer>>();
		for(int i = 0; i < M; i++) {
			if(!map.containsKey(x[i])) {
				ArrayList<Integer> list = new ArrayList<Integer>();
				list.add(y[i]);
				map.put(x[i], list);
			}else {
				map.get(x[i]).add(y[i]);
			}
			if(!map.containsKey(y[i])) {
				ArrayList<Integer> list = new ArrayList<Integer>();
				list.add(x[i]);
				map.put(y[i], list);
			}else {
				map.get(y[i]).add(x[i]);
			}
		}

//		for(int i : map.keySet()) {
//			System.out.print(i + " : ");
//			for(int j = 0; j < map.get(i).size(); j++) {
//				System.out.print(map.get(i).get(j) + " ");
//			}
//			System.out.println();
//		}

		int []color = new int[N + 1];
		Arrays.fill(color, 0);

		assignColor(color);
//		for(int i = 1; i <= N; i++) {
//			System.out.println(color[i]);
//		}

		int t = 0;
		int sum = 0;

		for(int i = 1; i <= N; i++) {
//			for(int j = 0; j < N; j++) {
//				if(i == p[j]) {
//					t = j + 1;
//					break;
//				}
//			}
			if(color[p[i - 1]] == color[i]) {
				sum ++;
			}
		}

		System.out.println(sum);


	}
	public static void dfs(int r, int c, int[] color) {
		Deque<Integer> stack = new ArrayDeque<Integer>();
		stack.push(r);
		color[r] = c;
		while(!stack.isEmpty()) {
			int u = stack.getFirst();
			stack.pop();
			if(map.containsKey(u)) {
				for(int i = 0; i < map.get(u).size(); i++) {
					int v = map.get(u).get(i);
					if(color[v] == 0) {
						color[v] = c;
						stack.push(v);
					}
				}
			}
		}
	}
	public static void assignColor(int[] color) {
		int id = 1;
		for(int i = 1; i <= N; i++) {
			if(color[i] == 0) {
				dfs(i, id++, color);
			}
		}
	}
}