Problem:
Score : 300 points

Problem Statement

There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i.
There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road.
Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq H_i \leq 10^9

1 \leq A_i,B_i \leq N

A_i \neq B_i

Multiple roads may connect the same pair of observatories.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Print the number of good observatories.

Sample Input 1

4 3
1 2 3 4
1 3
2 3
2 4

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use std::io::stdin;
use std::str::FromStr;
use std::collections::HashMap;

fn read_line<T>() -> Vec<T>
where
    T: FromStr,
    <T as std::str::FromStr>::Err: std::fmt::Debug,
{
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect()
}

fn main() {
    let in_vec: Vec<usize> = read_line();
    let n = in_vec[0];
    let m = in_vec[1];

    let h: Vec<usize> = read_line();

    let mut link_map: HashMap<usize, Vec<usize>> = HashMap::new();
    for _i in 0..m {
        let in_vec: Vec<usize> = read_line();
        let a = in_vec[0];
        let b = in_vec[1];
        if link_map.contains_key(&(a - 1)) {
            (*(link_map.get_mut(&(a - 1)).unwrap()))[b - 1] = 1;
        } else {
            link_map.insert(a - 1, vec![0; n]);
            (*(link_map.get_mut(&(a - 1)).unwrap()))[b - 1] = 1;
        }
        if link_map.contains_key(&(b - 1)) {
            (*(link_map.get_mut(&(b - 1)).unwrap()))[a - 1] = 1;
        } else {
            link_map.insert(b - 1, vec![0; n]);
            (*(link_map.get_mut(&(b - 1)).unwrap()))[a - 1] = 1;
        }
    }

    let mut res = 0;
    'outer: for key in 0..n {
        if link_map.contains_key(&key) {
            let key_hight = h[key];
            let val = link_map.get(&key).unwrap();
            for &val_i in val.iter() {
                if val_i == 0 {
                    continue;
                }
                let val_height = h[val_i];
                if key_hight <= val_height {
                    continue 'outer;
                }
            }
            res += 1;
        } else {
            res += 1;
        }
    }

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