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 {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
        for(int i = 1; i <= n; i++){
            map.put(sc.nextInt(),i);
        }
        UnionFindTree uft = new UnionFindTree(n+1);
        for(int i = 0; i < m; i++){
            uft.union(sc.nextInt(),sc.nextInt());
        }
        int ans = 0;
        for(int i = 1; i <= n; i++){
            if(uft.same(i,map.get(i))) ans++;
        }
        System.out.println(ans);
    }
}

class UnionFindTree {
	int[] parent;
	int[] rank;
	int[] size;
	
	public UnionFindTree(int size) {
		this.parent = new int[size];
		this.rank = new int[size];
		this.size = new int[size];
		
		for (int i = 0; i < size; i++) {
			makeSet(i);
		}
	}
	
	public void makeSet(int i) {
		parent[i] = i;
		rank[i] = 0;
		size[i] = 1;
	}
	
	public void union(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);
		
		if (rank[xRoot] > rank[yRoot]) { 
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
		} else if(rank[xRoot] < rank[yRoot]) {
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
		} else if (xRoot != yRoot){
			parent[yRoot] = xRoot;
			size[xRoot] += size[yRoot];
			rank[xRoot]++; 
		}
	}
	
	public int find(int i) {
		if (i != parent[i]) {
			parent[i] = find(parent[i]);
		}
		return parent[i];
	}
	
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}

	public int size(int x) {
		return size[find(x)];
	}
}