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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

int main(){
    int n, m;
    cin >> n >> m;

    int p[n];
    for(int i = 0; i < n; i++){
        cin >> p[i];
        p[i]--;
    }

    vector<vector<int> > v(n);
    for(int i = 0; i < m; i++){
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        v[x].push_back(y);
        v[y].push_back(x);
    }

    int ans = 0;
    for(int i = 0; i < n; i++){
        if(p[i] == i){
            ans++;
        }else{
            int flag[n];
            for(int j = 0; j < n; j++){
                flag[j] = 0;
            }
            if(v[i].size() != 0){
                queue<int> q;
                q.push(i);
                int check = 0;
                while(!q.empty()){
                    int now = q.front();
                    q.pop();
                    for(int j = 0; j < v[now].size(); j++){
                        if(p[v[now][j]] == i){
                            check = 1;
                            break;
                        }else{
                            if(flag[v[now][j]] == 0){
                                q.push(v[now][j]);
                            }
                        }
                    }
                    flag[now] = 1;
                    if(check == 1){
                        break;
                    }
                }

                if(check == 1){
                    ans++;
                }
            }
        }
    }

    cout << ans << endl;
}