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>
#include <vector>
#include <iostream>
#include<algorithm>
#include<string>
#include <map>
#include <queue>
#include <stack>
#include<set>

#define DIV 1000000007

using namespace std;
using ll = long long;
using ldb = long double;

typedef struct {
	int par = 0, hei = 1;
}node_t;
vector<node_t> node;

int root(int a) {
	if (node[a].par == 0) return a;
	node[a].par = root(node[a].par);
	return node[a].par;
}

void unite(int a, int b) {
	int art = root(a), brt = root(b);
	if (art == brt) return;
	int up, lo;
	if (node[art].hei >= node[brt].hei) up = art, lo = brt;
	else up = brt, lo = art;

	node[lo].par = up;
	node[up].hei = max(node[up].hei, node[lo].hei + 1);
}

int main() {
	int N, M; cin >> N >> M;
	vector<int> p(N + 1);
	for (int i = 1; i <= N; i++) {
		cin >> p[i];
	}
	node = vector<node_t>(N + 1);
	for (int i = 0; i < M; i++) {
		int x, y; cin >> x >> y;
		unite(x, y);
	}

	int ans = 0;
	for (int i = 1; i <= N; i++) {
		if (root(i) == root(p[i]))ans++;
	}
	cout << ans << endl;
}