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>

#define ll long long
#define pb push_back
#define eb emplace_back
//#define st first
#define sc second

using namespace std;

const ll mod = 1e9 + 7;
const int e = 1e5 + 69;
const int base = 311;

// dung inverse module de tinh nCk
// beginer 78
// mbeginer
// cach tao hoan vi tu n so : n*(n+1)/2;

//vector<pair<ll,ll>>adj[e];
vector<ll>adj[e];

/*ll Pow(ll n, ll dem)
{
    if(dem == 0) return 1;
    if(dem == 1) return n ;
    ll t = Pow(n, dem/2);
    if(dem % 2 == 0) return (t*t);
    else return ((t * t) * n);
}*/

/*ll C(ll k, ll n)
{
    if(k == 0) return 1;
    if(n < k) return 0;
    return (gt[n]%mod * (igt[k]%mod * igt[n-k]%mod)%mod)%mod;
}*/

int n, m, k;
ll h[e];

int main()
{
    cin >> n >> m;
    for(int i=1; i<=n; i++)
    {
        cin >> h[i];
    }
    map<ll,ll>kt;
    for(int i=1; i<=m; i++)
    {
        ll u, v;
        cin >> u >> v;
        //adj[u].pb(v);
        //adj[v].pb(u);
        if(h[u] > h[v]) kt[u] ++;
        else kt[v] ++;
    }
    ll dem = 0;
    for(int i=1; i<=n; i++)
    {
        if(kt[i] > 0) dem ++;
    }
    cout << dem;
}
/*

*/