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)

//I am so vegetable
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define FIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define PB push_back
#define IN insert
#define ER erase
#define CT count
#define EP empty
#define _for(i,a,b) for(int i = (a);i<(b);++i)
#define _rfor(i,a,b) for(int i = (a);i>(b);--i)
#define _sfor(p,st) for(auto &p:st)
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const double eps = 1e-6;
const double pi = 3.141593;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
ll qpow(ll x, ll y) { ll ans = 1; for (; y > 0; y >>= 1) { if (y & 1)ans *= x; x *= x; }return ans; }
ll qpow(ll x, ll y, int MOD) { ll ans = 1; for (; y > 0; y >>= 1) { if (y & 1)ans = ans*x%MOD; x = x*x%MOD; }return ans; }
void exgcd(ll a,ll b,ll &x,ll &y){
    if(!b){x=1;y=0;return;}
    exgcd(b,a%b,x,y);
    ll X=x;
    x=y; y=X-a/b*y;
}
ll inv(ll x, int mod = MOD) { return qpow(x, mod - 2) % mod; }
ll m_m(ll a){ return (a%MOD+MOD)%MOD; }

const int N = 1e5+5;
vector<int>edge[N];
int h[N];
int main() {
	//ofstream cout("E:\\acm dream\\out.txt",ios::out);
	FIO;
	int n,m;cin>>n>>m;
    _for(i,1,n+1)cin>>h[i];
    _for(i,0,m){
        int x,y;cin>>x>>y;
        edge[x].PB(y),edge[y].PB(x);
    }
    int sssss = 0;
    _for(i,1,n+1){
        int flag = 1;
        _sfor(it,edge[i]){
            if(h[it]>=h[i]){
                flag = 0;
                break;
            }
        }
        if(flag)++sssss;
    }
    cout<<sssss;
	return 0;
}