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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod=7+1e9;
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
ll qpow(ll a,ll b,ll p){a%=p; ll ret=1;for(;b;b>>=1,a=a*a%p) if(b&1) ret=ret*a%p; return ret;}
ll getInv(ll x,ll p){return qpow(x,p-2,p);}
const int N=5+1e5;
int h[N];
vector<int> G[N];
int n,m;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",h+i);
    for(int i=1;i<=m;i++){
        int u,v; scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    int cnt=0;
    for(int i=1;i<=n;i++){
        bool ok=1;
        for(auto v:G[i])
            if(h[i]<=h[v]){
                ok=0;
                break;
            }
        if(!ok) cnt++; 
    }
    printf("%d\n",cnt);
}