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::*;
use itertools::Itertools;
use proconio::{input, marker::*};
use superslice::*;
use whiteread::parse_line;
use std::collections::HashMap;

struct Obs {
    h: usize,
    is_toll: bool,
}

fn main() {
    // N 展望台の数 M ルートの数
    let (n, m): (usize, usize) = parse_line().unwrap();
    // H[N] 展望台の高さ
    let h_list: Vec<usize> = parse_line().unwrap();
    let mut obs_list: Vec<Obs> = vec![];

    // 構造体の初期化
    for idx in 0..n {
        obs_list.push(Obs{ 
            h: h_list[idx],
            is_toll: true,
        });
    }

    // 高さの算出
    for _ in 0..m {
        let (a, b): (usize, usize) = parse_line().unwrap();
        // 高さの解決
        fn calc (target: &Obs, comparison: &Obs) -> bool {
            if target.is_toll {
                return target.h > comparison.h
            }
            false
        };
        obs_list[a - 1].is_toll = calc(&obs_list[a - 1], &obs_list[b - 1]);
        obs_list[b - 1].is_toll = calc(&obs_list[b - 1], &obs_list[a - 1]);
    }

    let mut count = 0;
    for obs in obs_list {
        if obs.is_toll {
            count = count + 1;
        }
    }
    print!("{}", count);
}