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 proconio::{input, fastout};
use std::collections::{HashMap, HashSet};

#[fastout]
fn main() {
    input! {
        n: usize,
        m: usize,
        hx: [usize; n],
        ab: [(usize, usize); m]
    }
    
    let mut g: HashMap<usize, Vec<usize>> = HashMap::new();

    for i in 0..n {
        g.insert(i, vec![]);
    }

    for (a, b) in ab.into_iter() {
        let a = a-1;
        let b = b-1;
    
        (*g.entry(a).or_insert_with(|| vec![])).push(b);
        (*g.entry(b).or_insert_with(|| vec![])).push(a);
    }

    let mut high = vec![0;n];
    for i in 0..n {
        high[i] = i;
    }

    for (&i, v) in g.iter() {
        let mut highest = hx[i];
        let mut index = i;
        let mut same = false;
        for &j in v.iter() {
            if highest < hx[j] {
                highest = hx[j];
                index = j;
                same = false;
            } else if highest == hx[j] {
                same = true;
            }
        }
        high[i] = index;
        if same {
            high[i] = n;
        }
    }

    let mut hs = HashSet::new();

    for i in high.into_iter() {
        if i == n {
            continue;
        }
        hs.insert(i);
    }


    println!("{}", hs.len());

}