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::*;
use std::str::FromStr;
use std::cmp::Ordering;

fn main() {
	let (n, m): (usize, usize) = (read(), read());
	let mut h = Vec::new();
	let (mut a, mut b): (usize, usize);
	for _ in 0..n {
		a = read();
		h.push(a);
	}
	let mut v = vec![0; n];
	for _ in 0..m {
		a = read();
		b = read();
		match h[a-1].cmp(&h[b-1]) {
			Ordering::Less => { v[a-1] += 1; },
			Ordering::Greater => { v[b-1] += 1; },
			Ordering::Equal => { v[a-1] += 1; v[b-1] += 1; },
		}
	}
	let mut cnt: usize = 0;
	for i in &v {
		if *i == 0 { cnt += 1; }
	}
	println!("{}", cnt);
}

fn read<T: FromStr>() -> T {
	let stdin = stdin();
	let stdin = stdin.lock();
	let token: String = stdin
		.bytes()
		.map(|c| c.expect("failed to read char") as char)
		.skip_while(|c| c.is_whitespace())
		.take_while(|c| !c.is_whitespace())
		.collect();
	token.parse().ok().expect("failed to parse token")
}