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)

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            line.pop();
            line
        }
    )
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let iter = line.split_whitespace();
            iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    )
}

use std::rc::Rc;
use std::cell::RefCell;

struct UnionFind {
    par: Rc<RefCell<Vec<usize>>>
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let uf = UnionFind{ par: Rc::new(RefCell::new(vec![0; n])) };
        for i in 0..n {
            uf.par.borrow_mut()[i] = i;
        }
        uf
    }

    fn root(&self, x: usize) -> usize {
        let t = self.par.borrow()[x];
        if  t == x {
            x
        } else {
            self.par.borrow_mut()[x] = self.root(t);
            self.par.borrow()[x]
        }
    }

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

    fn unite(&self, mut x: usize, mut y: usize) {
        x = self.root(x);
        y = self.root(y);
        if x == y{ return }

        self.par.borrow_mut()[x] = y;
    }
}

fn solve(ps: Vec<usize>, xys: Vec<(usize, usize)>) -> usize {
    use std::collections::HashSet;

    let n = ps.len();
    let mut hs = HashSet::new();
    let uf = UnionFind::new(100001);

    for (x, y) in xys {
        uf.unite(x,y);
    }

    for i in  0..n {
        if ps[i] == i+1 {
            hs.insert(ps[i]);
        }

        if uf.root(ps[i]) == uf.root(i+1) {
            hs.insert(ps[i]);
        }
    }

    hs.len()
}

fn main() {
    let (_n, m) = parse_line!(usize, usize);

    let ps = parse_vec!(usize);

    let xys = (0..m).map(|_| parse_line!(usize, usize)).collect();

    println!("{}", solve(ps, xys));
}

#[test]
fn solve_test() {
    assert_eq!(solve(vec![5,3,1,4,2], vec![(1,3), (5,4)]), 2);
    assert_eq!(solve(vec![3,2,1], vec![(1,2), (2,3)]), 3);
    assert_eq!(solve(vec![5, 3, 6, 8, 7, 10, 9, 1, 2, 4], vec![(3,1), (4,1), (5,9), (2,5), (6,5), (3,5), (8,9), (7,9)]), 8);

}