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: Wrong Answer)

"use strict";

const main = arg => {
    const inputList = arg.trim().split("\n");
  // 回答を記述
  const N = parseInt(inputList[0].split(' ')[0]);
  const M = parseInt(inputList[0].split(' ')[1]);
  const observatorys = inputList[1].split(' ').map(ele => parseInt(ele));
  // 道のみのリストにする
  inputList.shift();
  inputList.shift();
  let roots = inputList;
  let result = 0;

  for(let i = 1; i <= N; i++) {
    // i番目の展望台と繋がる展望台道の組み合わせを抽出
    let matchRoots = roots.filter(ele => i == parseInt(ele.split(" ")[0]) || i == parseInt(ele.split(" ")[1]));
    // i番目の展望台とつながる展望台の組み合わせ
    let trimMatchRoots = matchRoots.map(ele => ele.split(" ")).flatMap(ele => ele).filter(ele => parseInt(ele) != i);
    let isHigh = true;
    for(let j = 0; j < trimMatchRoots.length; j++) {
      if (observatorys[i - 1] <= observatorys[parseInt(trimMatchRoots[j]) -1]) {
        isHigh = false;
        break;
      }
    }
    if (isHigh) {
      result++;
      //最高地点が含まれているものは除去する
      roots = roots.filter(ele => matchRoots.indexOf(ele) < 0)
    }
    
  }

  console.log(result);

}

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