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(unused_imports)]
#![allow(non_snake_case)]
use std::io::Read;
use std::io::{stdout, BufWriter, Write};
use std::cmp::min;

use std::collections::HashMap;
 
fn main() {
    let mut buf = String::new();
    
    let out = stdout();
    let mut out = BufWriter::new(out.lock());
 
    std::io::stdin().read_to_string(&mut buf).unwrap();
 
    let mut iter = buf.split_whitespace();
 
    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: usize = iter.next().unwrap().parse().unwrap();
    
    let h: Vec<u64> = (0..n).map(|_| {iter.next().unwrap().parse().unwrap()}).collect();
    let mut counts = vec![0u64; n];
    
    for _ in 0..m {
        let a: usize = iter.next().unwrap().parse().unwrap();
        let b: usize = iter.next().unwrap().parse().unwrap();
        let a = a - 1;
        let b = b - 1;
        if h[a] > h[b] {
            counts[b] += 1;
        } else if h[a] < h[b] {
            counts[a] += 1;
        } else {
            // h[a] == h[b]
            counts[a] += 1;
            counts[b] += 1;
        }
    }
    
    
    let ans : isize = counts.iter().map(|k| {
        if k == &0 {1} else {0}
        }).sum();
    
    println!("{}",ans);
    
}