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: Wrong Answer)

use proconio::input;
use std::cmp::Reverse;
use std::collections::BTreeSet;
use std::ops::Bound::{Included, Excluded};

fn solve(n: usize, m: usize, h: Vec<i32>, ab: Vec<(usize, usize)>) -> usize {
    let mut e = BTreeSet::new();
    for (a, b) in ab {
        e.insert((a - 1, b - 1));
        e.insert((b - 1, a - 1));
    }
    let mut hh = h.into_iter().enumerate().collect::<Vec<(usize, i32)>>();
    hh.sort_by_key(|&(_, x)| Reverse(x));
    let mut bad = vec![false; n];
    let mut ans = 0;
    for i in 0..n {
        let k = hh[n - 1 - i].0;
        if !bad[k] {
            ans += 1;
        }
        let mut y = Vec::new();
        for &(_, b) in e.range((Included(&(k, 0)), Excluded(&(k + 1, 0)))) {
            bad[b] = true;
            y.push(b);
        }
        for b in y {
            e.remove(&(k, b));
            e.remove(&(b, k));
        }
    }
    ans
}

fn main() {
    input! {
        n: usize,
        m: usize,
        h: [i32; n],
        ab: [(usize, usize); m],
    }
    let r = solve(n, m, h, ab);
    println!("{}", r);
}