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<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<functional>
#include<list>
#include<bitset>
#include<numeric>
#include<utility>
#define INF (int)1e9
#define MOD 1000000007
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<ll, ll> p;

struct unionfind {
	// 親ノードと木の高さ
	// rankは根のindexに対してのみ意味を持つ
	vector<int> par, rank;
	unionfind(int n) {
		par = vector<int>(n);
		rank = vector<int>(n, 0);
		for (int i = 0; i < n; i++) {
			par[i] = i;
		}
	}
	int root(int i) {
		if (par[i] == i) {
			return i;
		}
		return par[i] = root(par[i]);
	}
	bool same(int i, int j) {
		return root(i) == root(j);
	}
	void unite(int i, int j) {
		int ri = root(i);
		int rj = root(j);
		if (ri == rj) {
			return;
		}
		// 高さが高い方の木の根を低い方の根の親とする
		// 高さが同じ場合は最終的な親の高さがプラス1
		if (rank[ri] < rank[rj]) {
			par[ri] = rj;
			return;
		}
		par[rj] = ri;
		if (rank[ri] == rank[rj]) {
			rank[ri]++;
		}
	}
};
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m;
	cin >> n >> m;
	vector<int> p(n);
	for (int i = 0; i < n; i++) {
		cin >> p[i];
		p[i]--;
	}
	unionfind u = unionfind(n);
	int x, y;
	for (int i = 0; i < m; i++) {
		cin >> x >> y;
		x--;
		y--;
		u.unite(x, y);
	}
	int ans = 0;
	for (int i = 0; i < n; i++) {
		if (u.same(p[i], i)) {
			ans++;
		}
	}
	cout << ans;
}





// 解説AC