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 (1.15.1) (judge verdict: Accepted)

use std::io::{self, BufRead};
use std::collections::BTreeSet;

fn dfs(es: &Vec<Vec<usize>>, rest: &mut BTreeSet<usize>, visited: &mut BTreeSet<usize>, n: usize) {
  visited.insert(n);
  rest.remove(&n);
  for i in &es[n] {
    if visited.contains(i) {
      continue;
    }
    dfs(es, rest, visited, *i);
  }
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let (n, m) = (xs[0], xs[1]);
  let ps: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse::<usize>().unwrap() - 1).collect();
  let mut es: Vec<Vec<usize>> = vec![vec![]; n];
  for _ in 0..m {
    let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
    let (x, y) = (xs[0] - 1, xs[1] - 1);
    es[x].push(y);
    es[y].push(x);
  }
  let mut rest: BTreeSet<usize> = (0..n).collect();
  let mut count: usize = 0;
  while !rest.is_empty() {
    let n = *rest.iter().next().unwrap();
    let mut visited: BTreeSet<usize> = BTreeSet::new();
    dfs(&es, &mut rest, &mut visited, n);
    let vs: BTreeSet<usize> = visited.iter().map(|v| ps[*v]).collect();
    count += visited.iter().filter(|v| vs.contains(v)).count();
  }
  println!("{}", count);
}