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 __ <<' '<<
#define ___ <<' '
#define bash push_back
#define ALL(x) x.begin(),x.end()
//#define int long long

struct IoSetup {IoSetup() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout <<fixed<<setprecision(12);
    cerr <<fixed<<setprecision(12);
}}IoSetup;

using ll = long long;
using pii = pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int SMOD = 1000000007;
constexpr int NMOD = 998244353;
constexpr int dx[]={1,0,-1,0,1,1,-1,-1};
constexpr int dy[]={0,1,0,-1,-1,1,-1,1};

inline bool inside(int x,int y,int w,int h){return (x>=0 && y>=0 && x<w && y<h);}
template<class T>bool chmax(T &a, const T &b){return(a<b?a=b,true:false);}
template<class T>bool chmin(T &a, const T &b){return(a>b?a=b,true:false);}
inline int qp(int a,ll b,int mo){ll ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}

int c[100100];
int d[100100];
int h[100100];

signed main() {

    int n, m;
    
    cin >> n >> m;
    int a, b;
    for(int i = 1; i <= n; i++)
        cin >> h[i];

    for(int i = 0; i < m; i++) {
        cin >> a >> b;
        c[a]++;
        c[b]++;
        if(h[a] > h[b]) d[a]++;
        if(h[a] < h[b]) d[b]++;
    }

    int sum = 0;

    for(int i = 1; i <= n; i++) {
        if(c[i] == d[i]) sum++;
    }

    cout << sum << endl;
        
  
    return 0;
}