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(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::collections::HashMap;
// use std::collections::VecDeque;
// use std::cmp::max;
// use std::cmp::min;


fn main() {
    input! {
        N: usize,
        M: usize,
        Hs: [i64; N],
        mut ABs: [(usize, usize); M],
    }

    // 0-origin
    for (a, b) in ABs.iter_mut() {
        *a -= 1;
        *b -= 1;
    }

    let mut neibours_map: HashMap<usize, Vec<usize>> = HashMap::new();
    for i in 0..N {
        neibours_map.insert(i, Vec::new());
    }

    for &(a, b) in ABs.iter() {
        neibours_map.get_mut(&a).unwrap().push(b);
        neibours_map.get_mut(&b).unwrap().push(a);
    }

    let mut ans = 0;
    for i in 0..N {
        let vec: &Vec<usize> = neibours_map.get(&i).unwrap();
        let cnt = vec.iter().filter(|&&neibour| Hs[neibour] >= Hs[i]).count();
        if cnt == 0 {
            ans += 1;
        }
    }

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