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)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(a){return a?cin[cid++]:+cin[cid++];}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}

var myOut = main();
if(myOut !== undefined)console.log(myOut);

function main(){
  var [n,m] = nexts(2);
  var h = nexts(n);
  var q = xArray(1,n);
  for(var i = 0; i < m; i++){
    var [a,b] = nexts(2);
    a--,b--;
    if(h[a] >= h[b])q[b] = 0;
    if(h[b] >= h[a])q[a] = 0;
  }
  return q.reduce((a,b)=>a+b);
}