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: Wrong Answer)

#[allow(unused)]
use proconio::marker::*;
use proconio::{fastout, input};
#[allow(unused)]
use std::cmp::{max, min, Reverse};
#[allow(unused)]
use std::collections::{BinaryHeap, HashMap, HashSet, VecDeque};

#[fastout]
fn main() {
    input! {
        n: usize,
        m: usize,
        p: [Usize1; n],
        xy: [(Usize1, Usize1); m],
    }

    let mut uf = UnionFind::new(n);
    for &(x, y) in xy.iter() {
        uf.unite(x, y);
    }

    let mut hm = HashMap::new();
    for i in 0..n {
        let v = hm.entry(uf.find(i)).or_insert(vec![]);
        v.push(i);
    }

    let mut ans = 0;
    for (_, v) in hm.iter() {
        let mut w = vec![];
        for &i in v.iter() {
            w.push(p[i]);
        }
        w.sort();

        let mut cur = v[0];
        for &x in w.iter() {
            while x >= cur && cur <= v[v.len() - 1] {
                if x == cur {
                    ans += 1;
                    cur += 1;
                    break;
                }
                cur += 1;
            }
        }
    }

    println!("{}", ans);
}

struct UnionFind {
    parents: Vec<usize>,
    rank: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut vec = vec![0; n];
        for i in 0..n {
            vec[i] = i;
        }
        UnionFind {
            parents: vec,
            rank: vec![0; n],
        }
    }

    fn find(&self, x: usize) -> usize {
        if x == self.parents[x] {
            x
        } else {
            self.find(self.parents[x])
        }
    }

    fn unite(&mut self, x: usize, y: usize) {
        let px = self.find(x);
        let py = self.find(y);
        if px == py {
            return;
        }

        if self.rank[px] < self.rank[py] {
            self.parents[px] = py;
        } else {
            self.parents[py] = px;
            if self.rank[px] == self.rank[py] {
                self.rank[px] += 1;
            }
        }
    }

    fn same(&self, x: usize, y: usize) -> bool {
        self.find(x) == self.find(y)
    }
}