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>

using namespace std;

#define inf 1<<30
#define llong long long
#define usd unsigned
#define pqueue priority_queue
#define mp make_pair
#define pb push_back
#define fn fill_n
typedef pair<int, int> P;
typedef struct data{
	int now;
	int cost;
}data;
int nw[9] = {0, 1, 0, -1, -1, 1, 1, -1, 0};
int nh[9] = {1, 0, -1, 0, 1, 1, -1, -1, 0};

int n, m;
P p[100001], xy[100001];

int par[100001];

int root(int x){
	if(par[x] == x) return x;
	else return par[x] = root(par[x]);
}

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

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

	int dum;
	if(x < y){
		par[y] = x;
	}else{
		par[x] = y;
	}

	return;
}

int solve(){
	int ans = 0;
	for(int i = 1; i <= n; i++){
		par[i] = i;
	}

	sort(xy, xy+m);
	for(int i = 0; i < m; i++){
		unite(xy[i].first, xy[i].second);
	}
	for(int i = 1; i <= n; i++){
		if(same(p[i].first, p[i].second)) ans++;
	}

	return ans;
}

int main(){
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		cin >> p[i].first;
		p[i].second = i;
	}
	for(int i = 0; i < m; i++){
		cin >> xy[i].first >> xy[i].second;
		if(xy[i].first > xy[i].second){
			int dum = xy[i].first;
			xy[i].first = xy[i].second;
			xy[i].second = dum;
		}
	}

	cout << solve() << endl;

	return 0;
}