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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <climits>
#include <bitset>
#include <utility>
#include <cstring>
#include <numeric>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<pair<int,int>> vpii;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvc;
typedef vector<vector<string>> vvs;
#define rep(i,n) for(int i=0;i<n;i++)
#define repll(i,n) for(long long int i=0;i<n;i++)
#define fin(ans) cout<<(ans)<<endl
#define P 1000000007
#define STI(s) atoi(s.c_str())//string to int
#define mp(p,q) make_pair(p,q)
#define Sort(a) sort(a.begin(),a.end())
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
const int INF = INT_MAX;
const long long LLINF = 1LL<<60;
template <typename T_char>
char ToUpper (char cX) { return toupper(cX); }
//g++ -std=c++1z temp.cpp
//./a.exe
ll digit(ll a){
    ll count = 0;
    for (ll i = 0; i < a; i++)
    {   if (a == 0)
        {
            break;
        }else
        {
            a = a/10;
            count++;
        }
    }
    return count;
}

uintmax_t comb(unsigned int n, unsigned int r) {
  if ( r * 2 > n ) r = n - r;
  uintmax_t dividend = 1;
  uintmax_t divisor  = 1;
  for ( unsigned int i = 1; i <= r; ++i ) {
    dividend *= (n-i+1);
    divisor  *= i;
  }
  return dividend / divisor;
}
//桁数のカウント
//pow(x,n) == x^n
//sum = accumulate(v.begin(),v.end(),0);
using Graph = vector<vector<int>>;




int main(void){ios::sync_with_stdio(false);cin.tie(0);
/////////////////////////////////////////////////////

    int count = 0;

    // 頂点数と辺数
    int N, M; cin >> N >> M;
    vector<int> h(N);
    rep(i,N) cin >> h[i];

    // グラフ入力受取 (ここでは無向グラフを想定)
    Graph G(N);
    for (int i = 0; i < M; ++i) {
        int a, b;
        cin >> a >> b;
        G[a-1].push_back(b-1);
        G[b-1].push_back(a-1);
    }

    // 頂点 0 をスタートとした探索
    rep(i,N){
      bool check = true;
      int high = h[i];
      rep(j,G[i].size()){
        if(h[i] <= h[G[i][j]]) check = false;
      }
      if(check == true) count++;
    }
    cout << count << endl;


//////////////////////////////////////////////////////
return 0;}