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.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.ArrayList;

public class Main {
	Scanner sc = new Scanner(System.in);
	int n, m;
	int[] p; 
	Pair[] pairs;
	public void run(){
		n = sc.nextInt();
		m = sc.nextInt();
		p = new int[n];
		for(int i = 0; i < n; i++){
			p[i] = sc.nextInt() - 1;
		}
		pairs = new Pair[m];
		for(int i = 0; i < m; i++){
			int x = sc.nextInt() - 1;
			int y = sc.nextInt() - 1;
			
			if(x < y) pairs[i] = new Pair(x, y);
			else pairs[i] = new Pair(y, x);
		}
		Arrays.sort(pairs, new Comparator<Pair>(){
			public int compare(Pair o1, Pair o2){
				return Integer.compare(o1.first, o2.first);
			}
		});
		solve();
	}
	public void solve(){
		int[] list = new int[n];
		
		int gnum = 0;
		for(int i = 0; i < m; i++){
			Pair npair = pairs[i];
			int check1 = list[npair.first];
			int check2 = list[npair.second];
			
			int g = 0;
			
			if(check1 == 0 && check2 == 0){
				gnum = gnum + 1;
				g = gnum;
			}
			else if(check1 == 0){
				g = check2;
			}
			else if(check2 == 0){
				g = check1;
			}
			else{
				g = check1;
				for(int j = 0; j < n; j++){
					if(list[j] == check2){
						list[j] = check1;
					}
				}
			}
			
			list[npair.first] = g;
			list[npair.second] = g;
		}
		
		int ans = 0;
		
		for(int i = 0; i < n; i++){
			if(list[i] == list[p[i]]){
				ans = ans + 1;
			}
		}
		
		System.out.println(ans);
	}
	
	public static void main(String[] args){
		new Main().run();
	}
	
	static class Pair {
        int first,second;
        public Pair(int a,int b) {
            this.first = a;
            this.second = b;
        }
    }	
}