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 <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>
#include <map>
#include <set>

using namespace std;

class UnionFind {
private:
public:
	// parents[i] >= 0 なら i の親
	// parents[i] < 0 で、根。abs(parents[i])がその木に属する頂点数
	vector<int> parents;
	int n_root;

	UnionFind();
	UnionFind(int n) {
		parents.assign(n, -1);
		n_root = n;
	}
	~UnionFind() {
		parents.clear();
	};

	int find(int x) {
		if (parents[x] < 0) {
			return x;
		} else {
			return parents[x] = find(parents[x]);
		}
	}

	void unite(int x, int y) {
		x = find(x);
		y = find(y);

		if (x == y) {
			return;
		}

		if (size(x) < size(y)) {
			swap(x, y);
		}

		parents[x] += parents[y];
		parents[y] = x;
		n_root--;
		return ;
	}

	bool same(int x, int y) {
		return find(x) == find(y);
	}

	// xが属する集団のサイズ
	int size(int x) {
		return -parents[find(x)];
	}

	// いくつの集団があるか
	int get_n_root() {
		return n_root;
	}
};

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

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

	UnionFind uf(n + 1);
	for (int i = 0; i < m; i++) {
		cin >> x >> y;
		uf.unite(x, y);
	}

	map<int, set<int>> group_map;
	vector<int> parents;
	for (int i = 1; i <= n; i++) {
		int parent = uf.find(i);
		if (group_map.count(parent) == 0) {
			parents.emplace_back(parent);
		}
		group_map[parent].insert(i);
		// cout << i << " -> " << parent << endl;
	}

	int ans = 0;
	for (auto parent: parents) {
		auto s_gr = group_map[parent];
		vector<int> v_gr(s_gr.begin(), s_gr.end());

		// cout << parent << ": ";
		// for (auto a: v_gr) {
		// 	cout << a << " ";
		// }
		// cout << endl;

		for (auto z: v_gr) {
			if (s_gr.count(p[z - 1]) > 0) {
				ans++;
			}
		}
	}

	cout << ans << endl;
	return 0;
}