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 Rust (1.42.0) (judge verdict: Accepted)

use std::collections::BTreeSet;

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let p: Vec<usize> = s
		.by_ref()
		.take(n)
		.map(|a| a.parse::<usize>().unwrap() - 1)
		.collect();
	let mut edges = vec![vec![]; n];
	for _ in 0..m {
		let x: usize = s.next().unwrap().parse().unwrap();
		let y: usize = s.next().unwrap().parse().unwrap();
		edges[x - 1].push(y - 1);
		edges[y - 1].push(x - 1);
	}
	let mut visit = vec![false; n];
	let mut ans = 0;
	for i in 0..n {
		if visit[i] {
			continue;
		}
		let mut perm = BTreeSet::new();
		let mut pos = BTreeSet::new();
		dfs(&p, &edges, i, &mut visit, &mut perm, &mut pos);
		let int = perm.intersection(&pos).collect::<Vec<_>>();
		ans += int.len();
	}
	println!("{}", ans);
}

fn dfs(
	p: &Vec<usize>,
	edges: &Vec<Vec<usize>>,
	v: usize,
	visit: &mut Vec<bool>,
	perm: &mut BTreeSet<usize>,
	pos: &mut BTreeSet<usize>,
) {
	visit[v] = true;
	perm.insert(p[v]);
	pos.insert(v);
	for &to in edges[v].iter() {
		if !visit[to] {
			dfs(p, edges, to, visit, perm, pos);
		}
	}
}