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)

macro_rules! scan_sub {
    ( $itr:ident, $t:ty ) => {
        let mut buf = String::new();
        let $itr = {
            while buf.trim().len() == 0 {
                buf = String::new();
                std::io::stdin().read_line(&mut buf).unwrap();
            }
            buf.split_whitespace().map(|n| n.parse::<$t>().unwrap())
        };
    };
}

macro_rules! scan {
    ( $x:ident; Vec<$t:ty> ) => {
        scan_sub!(itr, $t);
        let $x = itr.collect::<Vec<_>>();
    };
    ( $( $x:ident ),*; $t:ty ) => {
        scan_sub!(itr, $t);
        let mut itr = itr;
        $(
            let $x = itr.next().unwrap();
        )*
    };
}

fn main() {
    scan!(n, m; usize);
    scan!(heights; Vec<usize>);

    let mut paths = vec![vec![0; 0]; n];
    for _ in 0..m {
        scan!(a, b; usize);
        paths[a - 1].push(b - 1);
        paths[b - 1].push(a - 1);
    }

    let res = paths.into_iter()
        .enumerate().filter(|(ip, p)| {
            let h = heights[*ip];
            p.iter().all(|&j| h > heights[j])
        }).collect::<Vec<_>>().len();

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