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)

#[allow(dead_code)]
fn main() {
    let (n, m) = i::u2();
    let h = i::uv(n);
    let ab = i::uv2(m);

    let mut s = vec![HashSet::new(); n + 1];
    for i in 0..m {
        let (a, b) = ab[i];
        s[a].insert(b);
        s[b].insert(a);
    }

    let mut count = 0;
    'case: for i in 1..=n {
        let height = h[i - 1];
        for &k in &s[i] {
            let target = h[k - 1];
            if height <= target {
                continue 'case;
            }
        }
        count += 1;
    }
    println!("{}", count);
}

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::*;
#[allow(unused_imports)]
use std::num::*;
#[allow(unused_imports)]
use std::str::*;

#[allow(dead_code)]
mod i {
    use super::*;

    pub fn read<T: FromStr>() -> T {
        stdin()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>()
            .parse::<T>()
            .ok()
            .unwrap()
    }

    pub fn str() -> String {
        read()
    }

    pub fn s() -> Vec<char> {
        str().chars().collect()
    }

    pub fn i() -> i64 {
        read()
    }

    pub fn u() -> usize {
        read()
    }

    pub fn u2() -> (usize, usize) {
        (read(), read())
    }

    pub fn u3() -> (usize, usize, usize) {
        (read(), read(), read())
    }

    pub fn f() -> f64 {
        read()
    }

    pub fn c() -> char {
        read::<String>().pop().unwrap()
    }

    pub fn iv(n: usize) -> Vec<i64> {
        (0..n).map(|_| i()).collect()
    }

    pub fn iv2(n: usize) -> Vec<(i64, i64)> {
        (0..n).map(|_| iv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv(n: usize) -> Vec<usize> {
        (0..n).map(|_| u()).collect()
    }

    pub fn uv2(n: usize) -> Vec<(usize, usize)> {
        (0..n).map(|_| uv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv3(n: usize) -> Vec<(usize, usize, usize)> {
        (0..n).map(|_| uv(3)).map(|a| (a[0], a[1], a[2])).collect()
    }

    pub fn fv(n: usize) -> Vec<f64> {
        (0..n).map(|_| f()).collect()
    }

    pub fn cmap(h: usize) -> Vec<Vec<char>> {
        (0..h).map(|_| s()).collect()
    }
}