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 <bits/stdc++.h>
#define endl '\n'
#define debug(X) cout << #X << " = " << X << endl

using namespace std;

const int MN = 100010;

int a[MN], par[MN];
set<int> dict[MN];

int findSet(int i) { return par[i] == i ? i : par[i] = findSet(par[i]); }
void unionSet(int i, int j) { par[findSet(i)] = findSet(j); }

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	#ifdef UTP
	freopen("input.txt", "r", stdin);
	#endif
	int n, m; cin >> n >> m;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	for (int i = 0; i < n; i++) par[i] = i;
	while (m--) {
		int x, y; cin >> x >> y; x--; y--;
		unionSet(x, y);
	}
	for (int i = 0; i < n; i++)
		dict[findSet(i)].insert(a[i] - 1);
	int ans = 0;
	for (int i = 0; i < n; i++)
		ans += dict[findSet(i)].count(i);
	cout << ans << endl;
	return 0;
}