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 itertools::Itertools as _;
use proconio::{input, marker::Usize1};

fn main() {
    input! {
        n: usize,
        m: usize,
        h: [i32; n],
        d: [[Usize1;2]; m],
    }

    let mut good_n = vec![true; n];

    for v in d.into_iter() {
        // eprintln!();
        // eprintln!("{:?}", h);
        // eprintln!("{:?}", v);
        // eprintln!("{:?}", good_n);
        if h[v[0]] >= h[v[1]] {
            good_n[v[1]] = false;
        }
        if h[v[0]] <= h[v[1]] {
            good_n[v[0]] = false;
        }
        // eprintln!("{:?}", good_n);
    }

    // //before
    // let before = d.iter().filter(|b| b[0] == now).map(|b| b[1]);
    // //after
    // let after = d.iter().filter(|a| a[1] == now).map(|a| a[0]);
    // //combine
    // let max = before
    //     .chain(after)
    //     .map(|target| h[target])
    //     .max()
    //     .unwrap_or(0);
    // if h[now] > max {
    //     ans += 1;
    // }

    println!("{}", good_n.into_iter().filter(|b| *b).count());
}