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;
typedef long double ld;
#define REP(i,n) for(int i=0; i<(n); i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define ALL(n) begin(n),end(n)
#define IN(a, x, b) (a<=x && x<b)
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
template<class T> inline T CHMAX(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T CHMIN(T & a, const T b) { return a = (a > b) ? b : a; }

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);

int main()
{
    INIT;
    int n, m;
    cin >> n >> m;
    vector<int> h(n);
    
    vector<int> copy(n);
    int sum = 0;
    REP(i,n){
        cin >> h.at(i);
        copy.at(i) = h.at(i);
    }

    int a,b;
    REP(i,m){
        cin >> a >> b;
        if(h.at(a-1) < h.at(b-1)){
            copy.at(a-1) = 0;
        }
        else if(h.at(a-1) > h.at(b-1)){
            copy.at(b-1) = 0;
        }else{
            if(copy.at(a-1) == 0) copy.at(b-1) = 0;
            if(copy.at(b-1) == 0) copy.at(a-1) = 0; 
        }
    }
    int ans = 0;
    REP(i,n){
        if(copy.at(i) != 0) ans++;
    }
    cout << ans << endl;
    return 0;
}