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)

#include <bits/stdc++.h>
using namespace std;

#define fa(x,v) for(auto x:v)
#define f(i,a,b) for(ll i=(ll)a;i<=(ll)b;i++)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i--)
#define spc " "
#define endl '\n'
#define ll long long
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define len(s) (ll)int(s.length())
#define sz(v) (ll)int(v.size())
#define MOD 1000000007

void Fast_IO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
    freopen("output.out","w",stdout);
    freopen("debug.out","w",stderr);
    #endif
} 

// Code Here

int main()
{
    // Fast_IO();
    ll n, m; cin >> n >> m;
    ll a[n]; f(i,0,n-1) cin >> a[i];
    vector < vector<ll> > adj;
    adj.resize(n);
    f(_,1,m) {
        ll u, v; cin >> u >> v;
        u--; v--;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    ll ans = 0;
    f(i,0,n-1) {
        ll flag = 1;
        f(j,0,sz(adj[i])-1) {
            ll v = adj[i][j];
            if(a[v] >= a[i]) flag = 0;
        }
        if(flag) ans++;
    }
    cout << ans;
    return 0;
}