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::cmp::Ordering::*;
use std::io;
use std::io::prelude::*;

fn main() -> Result<(), Box<dyn std::error::Error>> {
    let mut buffer = String::new();
    io::stdin().read_to_string(&mut buffer)?;
    let mut input = buffer.lines();
    input.next();
    let mut input = input.map(|s| s.split_whitespace());
    let mut elevations: Vec<(u32, bool)> = input.next().unwrap().map(|n| (n.parse().unwrap(), true)).collect();
    let input = input.map(|l| l.map(|n| n.parse::<usize>().unwrap() - 1));
    for mut pair in input {
        let left = pair.next().unwrap();
        let right = pair.next().unwrap();
        match elevations[left].0.cmp(&elevations[right].0) {
            Less => elevations[left].1 = false,
            Equal => { elevations[left].1 = false; elevations[right].1 = false; },
            Greater => elevations[right].1 = false,
        }
    }
    let output = elevations.into_iter().filter(|&(_, good)| good).count();
    println!("{}", output);
    Ok(())
}