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(dead_code)]
struct Utree {
    n: usize,
    par: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}
#[allow(dead_code)]
impl Utree {
    fn new(n: usize) -> Utree {
        let par: Vec<usize> = (0..n).collect();
        let rank: Vec<usize> = vec![0; n];
        let size: Vec<usize> = vec![1; n];
        Utree {
            n: n,
            par: par,
            rank: rank,
            size: size,
        }
    }
    fn find(&mut self, x: usize) -> usize {
        if self.par[x] == x {
            x
        } else {
            let parent = { self.par[x] };
            let ret = self.find(parent);
            self.par[x] = ret;
            ret
        }
    }
    fn unite(&mut self, x: usize, y: usize) {
        let x_root = self.find(x);
        let y_root = self.find(y);
        if x_root == y_root {
            return;
        };
        if self.rank[x_root] < self.rank[y_root] {
            self.par[x_root] = y_root;
            self.size[y_root] += self.size[x_root];
        } else {
            self.par[y_root] = x_root;
            self.size[x_root] += self.size[y_root];
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }
    fn is_same(&mut self, x: usize, y: usize) -> bool {
        let x_root = self.find(x);
        let y_root = self.find(y);
        x_root == y_root
    }
    fn get_size(&mut self, x: usize) -> usize {
        let x_root = self.find(x);
        self.size[x_root]
    }
}

fn main() {
    use std::io::Read;
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    
    let n: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let m: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let p: Vec<usize> = (&mut iter).take(n).map(|x| x.parse::<usize>().unwrap()-1).collect();
    let mut utree = Utree::new(n);
    for _ in 0..m {
        let x: usize = iter.next().unwrap().parse::<usize>().unwrap()-1;
        let y: usize = iter.next().unwrap().parse::<usize>().unwrap()-1;
        utree.unite(x, y);
    }

    let mut ret = 0usize;
    for i in 0..n {
        if utree.is_same(i, p[i]) {
            ret += 1;
        }
    }
    println!("{}", ret);
}