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: Accepted)

use std::io::stdin;

fn main() {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    let mut itr = s.split_whitespace().map(|e| e.parse().unwrap());
    let n: usize = itr.next().unwrap();
    let m: usize = itr.next().unwrap();

    s.clear();
    stdin().read_line(&mut s).ok();
    let h: Vec<usize> = s
        .split_whitespace()
        .map(|e| e.parse::<usize>().unwrap())
        .collect();
    s.clear();
    let mut edge: Vec<Vec<usize>> = vec![vec![]; 100009];

    for _ in 0..m {
        s.clear();
        stdin().read_line(&mut s).ok();
        let mut itr = s.split_whitespace().map(|e| e.parse::<usize>().unwrap());
        let a = itr.next().unwrap();
        let b = itr.next().unwrap();
        edge[a].push(b);
        edge[b].push(a);
    }

    let mut ans = 0;

    'outer: for i in 1..=n {
        let now = h[i - 1];
        for &e in &edge[i] {
            if now <= h[e - 1] {
                continue 'outer;
            }
        }

        ans += 1;
    }

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