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)

// Vicfred
// https://atcoder.jp/contests/abc166/tasks/abc166_c
use std::io;

fn main() {
    let mut line = String::new();

    io::stdin()
        .read_line(&mut line)
        .unwrap();

    let words: Vec<i64> =
        line
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();

    let n = words[0];
    let m = words[1];

    let mut line = String::new();

    io::stdin()
        .read_line(&mut line)
        .unwrap();

    let h: Vec<i64> =
        line
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();


    let mut graph = [0;100005];

    for _ in 0..m {
        let mut line = String::new();

        io::stdin()
            .read_line(&mut line)
            .unwrap();

        let words: Vec<usize> =
            line
            .split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect();

        let a = words[0]-1;
        let b = words[1]-1;

        if h[a] < h[b] {
            graph[a] = 1;
        } else if h[b] < h[a] {
            graph[b] = 1;
        } else {
            graph[a] = 1;
            graph[b] = 1;
        }
    }

    let mut answer = 0;
    for item in 0..n as usize {
        if graph[item] == 1 {
            answer += 1;
        }
    }

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