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 Java (OpenJDK 11.0.6) (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();
        int q[] = new int[N];
        for(int i = 0; i < N; i++) {
            int x = sc.nextInt() - 1;
            q[x] = i;
        }
        
        UnionFind nf = new UnionFind(N);
        
        for(int i = 0; i < M; i++) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            nf.unite(x, y);
        }
        
        int ans = 0;
        for(int i = 0; i < N; i++) {
            if( nf.isSame(q[i], i) ) {
                ans++;
            }
        }
        
        System.out.println(ans);
    }
}

class UnionFind {
    int[] parent;
    int[] size;
    int N;
    
    public UnionFind( int N ) {
        parent = new int[N];
        size = new int[N];
        this.N = N;
        for(int i = 0; i < N; i++) {
            parent[i] = i;
            size[i] = 1;
        }
    }
    
    public int root(int x) {
        if( parent[x] == x ) return x;
        return parent[x] = root(parent[x]);
    }
    
    public int getSize(int x) {
        return size[root(x)];
    }
    
    public boolean isSame(int x, int y) {
        return root(x) == root(y);
    }
    
    public void unite( int x, int y ) {
        if( isSame(x, y) ) return;
        size[root(y)] += size[root(x)];
        parent[root(x)] = root(y);
    }
    
}