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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;
typedef long long ll;

struct UnionFind
{
  vector<int> par; //par[i]:iの親の番号

  //最初は全てが根であるとする。-1のときに根である。
  UnionFind(int N) : par(N, -1) {}

  // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
  int root(int x)
  {
    if (par[x] < 0)
      return x;
    else
      return par[x] = root(par[x]);
  }

  //xとyが同じ集合に属するか
  bool issame(int x, int y)
  {
    return root(x) == root(y);
  }

  bool merge(int x, int y)
  {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    if (par[x] > par[y])
      swap(x, y);
    par[x] += par[y];
    par[y] = x;
    return true;
  }

  int size(int x)
  {
    return -par[root(x)];
  }
};

int main() {
  int n, m;
  int ans = 0;
  cin >> n >> m;
  vector<int> H(n);
  rep(i,n) cin >> H[i];
  vector<int> A[n];
  rep(i, m){
    int a, b;
    cin >> a >> b;
    a--; b--;
    A[a].emplace_back(b);
    A[b].emplace_back(a);
  }
  rep(i,n){
    bool cond = true;
    if(A[i].size() == 0){
      ans++;
    }else {
      int mx = H[i];
      rep(j,A[i].size()){
        if(mx <= H[A[i][j]]){
          cond = false;
        }
      }
      if(cond)ans++;
    }
  }
  cout << ans << endl;
}