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::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let (n , m) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };
    let p: Vec<usize> = read_vec();
    let mut union = UnionFind::new(n+1);
    for _ in 0..m {
      let (a, b) = {
        let t: Vec<i64> = read_vec();
        (t[0], t[1])
      };
      union.connect(a, b);
    }

    let mut ans = 0;
    for i in (1..n+1).rev() {
      if union.connected(i as i64, p[i - 1] as i64) {
        ans += 1;
      }
    }
    println!("{}", ans);
}



pub struct UnionFind {
  // 親のノードの番号を書くのする。親だった場合は-(その集合のサイズ)
  pub parent: Vec<i64>,
}


impl UnionFind {
  // コンストラクター
  pub fn new(n: usize) -> UnionFind {
    return UnionFind {parent: vec![-1; n]};
  }

  // aがどのグループに属しているか調べる
  // 属している木のの根を返す
  pub fn root(&mut self, a: i64) -> i64 {

    if self.parent[a as usize] < 0 {
      return a;
    }
    // aに属してる集合の根にaを直接つなぐ
    let tmp = self.parent[a as usize];
    self.parent[a as usize] = self.root(tmp);

    return self.parent[a as usize];
  }

  // 自分のいるグループの頂点数を調べる
  pub fn size(&mut self, a: i64) -> i64 {
    let tmp = self.root(a) as usize;
    return -self.parent[tmp];
  }

  // aとbをくっつける
  pub fn connect(&mut self, a: i64, b: i64) -> bool {
    let mut at = self.root(a);
    let mut bt = self.root(b);

    if at == bt {
      // すでにくっついてるからくっつけない
      return false;
    }

    // 大きい方aに小さい方bをくっつける
    // 大小が逆だったらひっくり返す
    if self.size(at) < self.size(bt) {
      let temp = at;
      at = bt;
      bt = temp;
    }

    // aのサイズを更新する
    self.parent[at as usize] += self.parent[bt as usize];
    // bの親をaに更新する
    self.parent[bt as usize] = at;
    return true
  }

  pub fn equal(&self, that: UnionFind) -> bool {
    let l: usize = that.parent.len();

    if self.parent.len() != l {
      println!("{}",l);
      println!("{:?}", self.parent);
      return false;
    }

    for i in 0..l {
      if self.parent[i] != that.parent[i] {
        return false;
      }
    }
    return true;
  }

  pub fn connected(&mut self, a: i64, b: i64 ) -> bool {
    self.root(a) == self.root(b)
  }
}