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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function main(input) {
  const inputs = input.trim().split('\n')
  const [N, M] = inputs[0].split(' ')
  const H = [0, ...inputs[1].split(' ')].map(Number)
  const path = []
  inputs.slice(2).forEach(e => {
    // console.log('e', e)
    const [A, B] = e.split(' ')
    if (!path[A]) {
      path[A] = []
    }
    path[A].push(B)
    if (!path[B]) {
      path[B] = []
    }
    path[B].push(A)
  })
  // console.log(path)

  let good = 0
  for(let i = 1; i <= N; i++) {
    let isGood = true
    if (!path[i]) {
      good++
      continue
    }
    for(const e of path[i]) {
      // console.log(i, H[i], e, H[e])
      if (H[e] >= H[i]) {
        isGood = false
        // console.log('not good')
        break
      }
    }
    if (isGood) good++
  }
  console.log(good)
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'))