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<iostream>
#include<vector>
#include<algorithm>
#include<stdlib.h>
#include<utility>
#include<functional>
#include<cfenv>
#include<cmath>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define vint vector<int>
#define vvint vector<vint>
#define P pair<int,int>
#define INT_MAX 2147483647
#define MOD 1000000007
using namespace std;
typedef long long ll;



int main(void) {
	int n, m; cin >> n >> m;
	vint h(n);
	vint f(n);
	rep(i, n) {
		cin >> h[i];
	}
	vvint e(n);
	rep(i, m) {
		int x,y;
		cin >>x >>y;
		x--; y--;
		e[x].push_back(y);
		e[y].push_back(x);
	}
	int ans = 0;
	rep(i, n) {
		if (f[i] == 0) {
			int j;
			for (j = 0; j < e[i].size();j++) {
				if (h[i] <= e[i][j]) {
					break;
				}
				else {
					f[e[i][j]] = 1;
				}
			}
			if (j == e[i].size()) {
				ans++;
			}
		}


	}
	cout << ans << endl;
	
	return 0;
}