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: Accepted)

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

#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef array<int, 3> tri;
typedef long double ld;

template <class T> istream& operator>>(istream& I, vector<T>& v) {for (T &e: v) I >> e; return I;}
template <class T> ostream& operator<<(ostream &O, const vector<T>& v) {for (const T &e: v) O << e << ' '; return O;}


void dfs(int a, const vector<vector<int>>& adj, vector<bool>& vis, vector<int>& cc) {
        cc.emplace_back(a);
        vis[a] = 1;
        for (auto b: adj[a]) {
                if (!vis[b]) {
                        dfs(b, adj, vis, cc);
                }
        }
}


void _main() {
        int n, m; cin >> n >> m;
        vector<int> perm(n + 1);
        for (int i = 1; i <= n; i++) cin >> perm[i];
        vector<vector<int>> adj(n + 1);
        while (m--) {
                int a, b; cin >> a >> b;
                adj[a].emplace_back(b);
                adj[b].emplace_back(a);
        }

        vector<bool> vis(n + 1);
        int ans = 0;
        for (int i = 1; i <= n; i++) {
                if (vis[i]) continue;
                vector<int> cc;
                dfs(i, adj, vis, cc);
                vector<int> bb;
                for (auto it: cc) bb.emplace_back(perm[it]);
                sort(all(bb));
                sort(all(cc));
                for (auto b: bb) {
                        ans += binary_search(all(cc), b);
                }
        }


        cout << ans;
}



signed main() {
        ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//        freopen("input.txt", "r", stdin);
        int _t = 1;
//        cin >> _t;
        while (_t--) _main();
        return 0;
}