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 proconio::input;
use proconio::marker::Chars;
use std::collections::HashSet;
use std::iter::Iterator;

fn main() {
    input! {
    n: i32,
    m: i32,
    h:[i32; n],
    mut road:[(i32, i32);m],
    }

    let mut ans = n;

    let mut v: HashSet<i32> = HashSet::new();

    for r in &road {
        if  h[(r.0 - 1) as usize] == h[(r.1 - 1) as usize]  {
            if !v.contains(&r.0) {
                v.insert(r.0);
                ans = ans -1;
            }
            if !v.contains(&r.1) {
                v.insert(r.1);
                ans = ans - 1;
            }
            continue;
        }

        if h[(r.0 - 1) as usize] < h[(r.1 - 1) as usize]  {
            if !v.contains(&r.0) {
                v.insert(r.0);
                ans = ans - 1;
            }
        } else {
            if !v.contains(&r.1) {
                v.insert(r.1);
                ans = ans - 1;
            }
        }
    }

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