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 <algorithm>

using namespace std;

long long int parr[100010];
long long int rankk[100010];

void init(long long int n) {
	for (long long int i = 0; i < n; i++) {
		parr[i] = i;
		rankk[i] = 0;
	}
}

long long int find(long long int x) {
	if (parr[x] == x) {
		return x;
	}
	else {
		return parr[x] = find(parr[x]);
	}
}

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

	if (rankk[x] < rankk[y]) {
		parr[x] = y;
	}
	else {
		parr[y] = x;
		if (rankk[x] == rankk[y]) {
			rankk[x]++;
		}
	}
}

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

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

	for (int i = 0; i <= n; i++) {
		find(i);
		//cout << i << ' ' << parr[i] << endl;
	}

	vector< vector<long long int> > aa(n+1);
	vector< vector<long long int> > bb(n+1);

	for (long long int i = 0; i <= n; i++) {
		aa[parr[i + 1]].push_back(i+1);
		bb[parr[i + 1]].push_back(p[i]);
	}

	//cout << 'a' << endl;

	long long int ans = 0;

	for (long long int i = 0; i < n+1; i++) {
		//cout << aa[i].size() << ' ' << bb[i].size() << 'a' << endl;
		sort(aa[i].begin(), aa[i].end());
		sort(bb[i].begin(), bb[i].end());
		long long int k = 0;
		for (long long int j = 0; j < aa[i].size(); j++) {
			//cout << aa[i][j] << ' ' << bb[i][k] << endl;
			if (k == bb[i].size()) {
				break;
			}
			if (aa[i][j] == bb[i][k]) {
				//cout << aa[i][j] << endl;
				ans++;
				k++;
			}
			else if (aa[i][j] < bb[i][k]) {

			}
			else {
				j--;
				k++;
			}
		}

	}

	cout << ans << endl;

	return 0;
}