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.*;
class Main
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        int[] a = new int[n], x = new int[m], y = new int[m];
        for(int i = 0; i < n; i++) a[i] =sc.nextInt()-1;
        UnionFindTree uft = new UnionFindTree(n);
        for(int i = 0; i < m; i++){
            uft.union(sc.nextInt()-1, sc.nextInt()-1);
        }
        int ans = 0;
        for(int i = 0; i < n; i++){
            if(uft.same(i, a[i])) ans++;
        }
        System.out.println(ans);
    }
}
class UnionFindTree
{
    int[] parent, rank;
    public UnionFindTree(int size)
    {
        this.parent= new int[size];
        this.rank = new int[size];
        for(int i = 0; i < size; i++){
            parent[i] = i;
            rank[i] = 0;
        }
    }
    public void union(int x, int y){
        int xRoot = find(x);
        int yRoot = find(y);

        if(rank[xRoot] > rank[yRoot]) parent[yRoot] = xRoot;

        else if(rank[xRoot] < rank[yRoot]) parent[xRoot] = yRoot;

        else if(xRoot != yRoot){
            parent[yRoot] = xRoot;
            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);
    }
}