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)

var obj = init();
Main();
function nextInt(){return myconv(next(),1);}
function nextStrArray(){return myconv(next(),2);}//半角スペース分割
function nextIntArray(){return myconv(next(),4);}//半角スペース + 数値化
function nextCharArray(){return myconv(next(),6);}//1文字分割
function next(){return obj.next();}
function hasNext(){return obj.hasNext();}
function myin(){return require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");}
function init(){
  var inputFile = myin();
  var returnObj = {
    "list" : inputFile, "index" : 0, "max" : inputFile.length,
    "hasNext" : function(){return (this.index < this.max);},
    "next" : function(){if(!this.hasNext()){throw "ArrayIndexOutOfBoundsException これ以上ないよ";}else{var returnInput = this.list[this.index];this.index++;return returnInput;}}
  };
  return returnObj;
}
function myout(t){console.log(t);}//標準出力
function myerr(t){console.error(t);}//標準エラー出力
//[no]要素の扱い。数値型
//不明値、異常時:引数そのまま返す　1:数値へ変換
//2:半角SPで分割 4:半角SPで分割し、数値配列へ
//6:1文字で分割  7:1文字で分割し、数値配列へ
//8:半角SPで結合 9:改行で結合 0:文字なしで結合
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map(Number);case 6:return i.split("");case 7:return i.split("").map(Number);case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}
function Main(){
  	var one = nextIntArray();
	var N = one[0];
	var M = one[1];
	var hlist = nextIntArray();
	var link = new Array(N);
	for(var i = 0; i < N; i++){
		link[i] = [];
	}
	var maxs = new Set();
	for(var i = 0; i < M; i++){
		var tmp = nextIntArray();
		link[tmp[0] - 1].push(tmp[1] - 1);
		link[tmp[1] - 1].push(tmp[0] - 1);
		if(hlist[tmp[0] - 1] > hlist[tmp[1] - 1]){
			maxs.add(tmp[0] - 1);
			maxs.delete(tmp[1] - 1);
		}else if(hlist[tmp[1] - 1] > hlist[tmp[0] - 1]){
			maxs.add(tmp[1] - 1);
			maxs.delete(tmp[0] - 1);
		}else{
			maxs.delete(tmp[1] - 1);
			maxs.delete(tmp[0] - 1);
		}
	}
	for(var i = 0; i < N; i++){
		if(link[i].length == 0){
			maxs.add(i);
		}
	}
	myout(maxs.size);
}