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)

/**
 * Dont raise your voice, improve your argument.
 * --Desmond Tutu
 */

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

vector<string> vec_splitter(string s) {
    for(char& c: s) c = c == ','?  ' ': c;
    stringstream ss; ss << s;
    vector<string> res;
    for(string z; ss >> z; res.push_back(z));
    return res;
}

void debug_out(vector<string> __attribute__ ((unused)) args, __attribute__ ((unused)) int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
    if(idx > 0) cerr << ", ";
    stringstream ss; ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, T...);
}

#ifdef XOX
#define debug(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __VA_ARGS__)
#else
#define debug(...) 42
#endif

const double PI = acos(-1);
typedef long long ll;
#define int ll
#define fori(n) for(int i=0; i<int(n); i++)

#define cins(s) string s; cin>>s;
#define cini(i) int i; cin>>i;
#define cind(d) double d; cin>>d;
#define cinai(a, n) vi a(n); fori(n) { cin>>a[i]; }
#define cinas(s, n) vs s(n); fori(n) { cin>>s[i]; }
#define cinad(a, n) vd a(n); fori(n) { cin>>a[i]; }

typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<string> vs;

void solve() {
    cini(n);
    cini(m);
    cinai(h,n);
    vvi adj(n);
    for(int i=0; i<m; i++) {
        cini(a);
        cini(b);
        a--;
        b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    vb ans(n, true);

    for(int i=0; i<n; i++) {
        for(int j : adj[i])
            if(h[i]<=h[j])
                ans[i]=false;
    }

    int cnt=0;
    for(int i=0; i<n; i++)
        if(ans[i])
            cnt++;
    cout<<cnt<<endl;
    
}

signed main() {
    cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    solve();
}