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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

int comp_id = 0;
vector<int> edges[100001];

int comp[100001];

void bfs(int from) {
    queue<int> q;
    q.push(from);
    comp[from] = comp_id;

    while (!q.empty()) {
        int current = q.front();
        q.pop();

        for (int to : edges[current]) {
            if (comp[to] == -1) {
                comp[to] = comp_id;
                q.push(to);
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int N, M;
    cin >> N >> M;

    vector<int> P;
    P.resize(N);

    int pos[100001];

    for (int i=0; i<N; i++) {
        cin >> P[i];
        pos[P[i]] = i;
    }

    for (int i=0; i<M; i++) {
        int f, t;
        cin >> f >> t;

        edges[f].push_back(t);
        edges[t].push_back(f);
    }

    memset(comp, -1, sizeof(comp));

    for (int i=1; i <= N; i++) {
        if (comp[i] == -1) {
            comp_id++;
            bfs(i);
        }
    }


    int result = 0;
    for (int i=0; i<N; i++) {
        int value = P[i];
        int actual_pos = pos[P[i]];
        int value_there = i + 1;

        if (comp[value] == comp[value_there]) {
            result++;
        }
    }

    cout << result << endl;
    return 0;
}