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)

#![allow(non_snake_case)]

use std::collections::{HashMap, HashSet, VecDeque};
use std::io::{stdin, Read};
use std::str::FromStr;
fn read_option<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok()
}
fn read<T: FromStr>() -> T {
    let opt = read_option();
    opt.expect("failed to parse token")
}

fn main() {
    let N: usize = read();
    let M: usize = read();
    let ps: Vec<usize> = (0..N).map(|_| read::<usize>() - 1).collect();
    let ms: Vec<(usize, usize)> = (0..M)
        .map(|_| (read::<usize>() - 1, read::<usize>() - 1))
        .collect();
    println!("{}", solve(ps, ms) + 1)
}

fn solve(ps: Vec<usize>, ms: Vec<(usize, usize)>) -> usize {
    // 1. msがつなぐグループを作る(i と値両方)
    let mut graph: HashMap<usize, Vec<usize>> = HashMap::new();
    for (x, y) in ms {
        let entry = graph.entry(x).or_insert_with(|| vec![]);
        entry.push(y);
        let entry = graph.entry(y).or_insert_with(|| vec![]);
        entry.push(x);
    }
    let mut visited = vec![false; ps.len()];
    let mut ans = 0;
    for i in 0..ps.len() {
        if visited[i] {
            continue;
        }
        let mut queue = VecDeque::new();
        queue.push_front(i);
        visited[i] = true;
        let mut indexes = HashSet::new();
        let mut values = HashSet::new();
        indexes.insert(i);
        values.insert(ps[i]);
        while let Some(v) = queue.pop_back() {
            match graph.get(&v) {
                None => {
                    continue;
                }
                Some(vec) => {
                    for &next in vec {
                        if visited[next] {
                            continue;
                        }
                        queue.push_front(next);
                        visited[next] = true;
                        indexes.insert(next);
                        values.insert(ps[next]);
                    }
                }
            }
        }
        for i in indexes {
            if values.contains(&i) {
                ans += 1;
            }
        }
    }
    ans - 1
}