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::{
   fastout, input,
   marker::{Bytes, Chars, Isize1, Usize1},
};

//use union_find::*;

fn main() {
   input! {
      n:usize,m:usize,
      mut h:[usize;n],
      ab:[(Usize1,Usize1);m]
   }
   let q = h.clone();
   let mut v = vec![vec![]; n];
   for (i, j) in ab {
      v[i].push(j);
      v[j].push(i);
   }
   let mut arr = vec![];
   let mut fg = vec![false; n];
   for i in 0..n {
      arr.push((h[i], i));
   }
   arr.sort();
   arr.reverse();
   let mut ans = 0_usize;
   for i in 0..n {
      if !fg[arr[i].1] {
         ans += 1;
         let mut flag = false;
         for j in &v[arr[i].1] {
            if q[arr[i].1] == q[*j] {
               flag = true;
            }
            fg[*j] = true;
         }
         if flag {
            ans -= 1;
         }
         fg[arr[i].1] = true
         //println!("{}", arr[i].1);
      }
      //println!("{:?}", &h);
   }
   println!("{}", ans);
}