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! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!{
        n: usize,
        m: usize,
        hh: [u32; n],
        ab: [(usize1, usize1); m],
    };

    let mut is_good = vec![true; n];
    for (a, b) in ab {
        let ha = hh[a];
        let hb = hh[b];
        if ha <= hb {
            is_good[a] = false;
        }
        if hb <= ha {
            is_good[b] = false;
        }
    }
    println!("{}", is_good.iter().fold(0, |a, b| a + if *b {1} else {0}));
}