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 C++ (GCC 9.2.1) (judge verdict: Accepted)

/*Author - Shubham Gupta (@shubham107)*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<ll> vl;		typedef vector<int> vi;
typedef vector<vl> vvl;	  	typedef vector<vi> vvi;
typedef pair<ll, ll> pll;	typedef pair<int, int> pii;
#define forn(i,n) for(int i=0;i<n;i++)
#define rforn(i,n) for(int i=n-1;i>=0;i--)
#define forne(i,n) for(int i=1;i<=n;i++)
#define forse(i,s,e) for(int i=s;i<e;i++)
#define rforse(i,s,e) for(int i=e-1;i>=s;i--)
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define MOD 1000000007
#define F first
#define S second
#define pb push_back
#define fill(a,x) memset(a,x,sizeof a);
#define trav(a,x) for(auto &a:x)
#define INF	ll(1e18)
#ifdef SHUBHAM107
#include "../code-pieces/trace.h"
#else
#define trace(args...)
#endif

ll modpow(ll a, ll b, ll m = MOD) {
	a %= m; ll res = 1; while (b) {if (b & 1)res = res * a % m; a = a * a % m; b >>= 1;}
	return res;
}
ll bpow(ll a, ll b) {ll res = 1; while (b) {if (b & 1)res = res * a; a = a * a; b >>= 1;} return res;}
ll modinv(ll a) {return modpow(a, MOD - 2, MOD);}
void graph(vvi &adj, int m) {int x, y; forn(i, m) {cin >> x >> y; adj[x - 1].pb(y - 1); adj[y - 1].pb(x - 1);}}

const int mxN = 2e5 + 5;
void solve() {
	int n, m;
	cin >> n >> m;
	vi heights(n);
	forn(i, n) cin >> heights[i];
	vvi adj(n);
	graph(adj, m);
	int ans = 0;
	forn(i, n){
		bool good = true;
		trav(k, adj[i]){
			if(heights[i] <= heights[k]) good = false;
		}
		if(good) ans++;
	}
	cout << ans;
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
#ifdef SHUBHAM107
	freopen("inp.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	freopen("err.txt", "w", stderr);
#endif
	int t = 1;
	// cin >> t;
	while (t--) {
		solve();
	}
}