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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define rep_bit(n) for (int bit = 0; bit < (1<<n); ++bit)
#define all(x) (x).begin(),(x).end()
#define ll long long
#define ld long double
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define SP <<" "<<
#define CST(x) cout<<fixed<<setprecision(x)
using namespace std;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vi> Graph;

//条件式 ? 真の場合の文 : 偽の場合の文
//abs(x) = (x の絶対値)
//string S  S,at(i),S += T,S += c,S.size(),S.substr(l)l文字目から最後まで,
///S.substr(l, r)l文字目からl+r-1文字目 n文字目から0個で空列を返せる
//*min_element(all(x)) *max_element(all(x))
//vi x; x.back(),x.front(),x.push_back(),x.pop_back()
//swap(a, b)
//gcd(a, b)最大公約数
//sort(all(x)),reverse(all(x)),sort(x.begin()+l,a.begin()+r)lからrまで
//stack<int> a; a.push(x),a.pop(),a.top(),a.empty()bool
//queue<int> b; b.push(x),b.pop(),b.front(),b.empty()bool,b.size() 要素が小さい順
//priority_queue<int, vector<int>, greater<int>> c,c.top()
//map<int, int> M1; 変数[key] = value; 変数.erase(key); 変数.at(key); 変数.count(key)所属判定　変数.size()
//find(all(x), key)boolソート済みのベクトル
//*lower_bound(all(x),k)以上の最小の要素,*uppper_bound(all(x),k)より大きい最小の要素
//char('0'+cnt) 数字cntからcharへの変換
//int(char-'0') charからintへの変換

int main(){
    int n,m;
    cin >> n >> m;
    vi h(n);
    rep(i,n){
        cin >> h.at(i);
    }

    vb ok(n,true);
    rep(i,m){
        int a,b;
        cin >> a >> b;
        a--;b--;
        if(!ok.at(i)) continue;
        if(h.at(a) <= h.at(b)) ok.at(a) = false;
        if(h.at(a) >= h.at(b)) ok.at(b) = false;
    }
    cout << count(all(ok),true) << endl;

}