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)

#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds; 

#define int long long
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define f first
#define s second
#define sz(x) (int)x.size()
#define endl '\n'
#define forn(i,n) for(int i=0;i<n;++i)
#define fore(i,l,r) for(int i=int(l);i<=int(r);++i)
#define rep(i,begin,end) for(__typeof(end) i=(begin);i!=(end);i++)
#define fill(a,value) memset(a,value,sizeof(a));
#define gcd(a,b) __gcd((a),(b))
#define watch1(x) cout << (x) << endl
#define watch2(x,y) cout << (x) << " " << (y) << endl
#define watch3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef long double ld; 
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> oset;

// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

const int mod = 1e9 + 7;
const int inf = 9e18;
const int N = 1e5 + 5;

int i,n,m,h[N];
vi g[N];

void solve(){
    cin>>n>>m;
    for(i=1;i<=n;++i){
        cin>>h[i];
    }
    for(i=0;i<m;++i){
        int a,b;
        cin>>a>>b;
        g[a].pb(b);
        g[b].pb(a);
    }
    int ans=0;
    for(i=1;i<=n;++i){
        int ok=1;
        for(int j:g[i]){
            if(h[i]<=h[j]){
                ok=0;
                break;
            }
        }
        ans+=ok;
    }
    cout<<ans;
}

signed main(){
    fastio;
    int t;
    //cin>>t;
    t=1;
    while(t--){
        solve();
        cout<<endl;
    }
    return 0;
}