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)]
use std::cmp;
use std::collections::*;

pub struct Scanner {
    line: Vec<String>,
    idx: usize,
}

#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner {
        return Scanner {
            line: vec![],
            idx: 0,
        };
    }
    fn next<T: std::str::FromStr>(&mut self) -> T {
        if self.idx == self.line.len() {
            let mut buffer = String::new();
            std::io::stdin().read_line(&mut buffer).ok();
            let a: Vec<&str> = buffer.trim().split(' ').collect();
            self.line = a.iter().map(|&s| String::from(s)).collect();
            self.idx = 0;
        }
        let ret: T = self.line[self.idx].parse::<T>().ok().expect("Parse error.");
        self.idx += 1;
        ret
    }
    fn next_string(&mut self) -> Vec<char> {
        let s: String = self.next();
        return s.chars().collect::<Vec<char>>();
    }
    fn next_vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        let mut v: Vec<T> = Vec::with_capacity(n);
        for _i in 0..n {
            v.push(self.next());
        }
        v
    }
}

fn main() {
    let mut scan = Scanner::new();
    let n: usize = scan.next();
    let m: usize = scan.next();
    let h: Vec<i32> = scan.next_vec(n);
    let mut ok: Vec<i32> = vec![1; n];

    for _ in 0..m {
        let a: usize = scan.next();
        let b: usize = scan.next();
        let (a, b) = (a - 1, b - 1);

        if h[a] <= h[b] {
            ok[a] = 0;
        }
        if h[a] >= h[b] {
            ok[b] = 0;
        }
    }
    println!("{}", ok.iter().fold(0, |sum, x| sum + x));
}