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::{self, BufRead};

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let (n, m) = (xs[0], xs[1]);
  let hs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let mut es: Vec<Vec<usize>> = vec![vec![]; n];
  for _ in 0..m {
    let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
    let (a, b) = (xs[0] - 1, xs[1] - 1);
    es[a].push(b);
    es[b].push(a);
  }
  let mut count: usize = 0;
  for i in 0..n {
    let h = hs[i];
    let mut b = true;
    for j in &es[i] {
      if h <= hs[*j] {
        b = false;
      }
    }
    if b {
      count += 1;
    }
  }
  println!("{}", count);
}