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: Time Limit Exceeded)

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        
        int[] par = new int[n];
        int[] rank = new int[n];
        int[] p = new int[n];
        for(int i=0; i<n; i++){
            p[i] = Integer.parseInt(sc.next());
            par[i] = i;
        }
        
        for(int i=0; i<m; i++){
            int x = Integer.parseInt(sc.next())-1;
            int y = Integer.parseInt(sc.next())-1;
            unite(par, rank, x, y);
        }
        
        List<Integer>[] h = new ArrayList[n];
        for(int i=0; i<n; i++){
            h[i] = new ArrayList<>();
        }
        
        for(int i=0; i<n; i++){
            int tmp = find(par, i);
            h[tmp].add(i);
            //System.out.println(tmp);
        }
        
        int ans = 0;
        for(int i=0; i<n; i++){
            for(int t : h[par[i]]){
                if(p[i] == t+1){
                    ans++;
                }
            }
        }
        
        System.out.println(ans);
    }
    
    public static int find(int[] par, int x){
        if(par[x] == x){
            return x;
        }else{
            par[x] = find(par, par[x]);
            return par[x];
        }
    }
    
    public static void unite(int[] par, int[]rank, int x, int y){
        x = find(par, x);
        y = find(par, y);
        if(x==y) return;
        
        if(rank[x]<rank[y]){
            par[x] = y;
        }else{
            par[y] = x;
            if(rank[x]==rank[y]) rank[x]++;
        }
    }
}