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)

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include <unordered_map>
//#include"testlib.h"
#define endl "\n"
#define all(v) v.begin(),v.end()
#define allr(s) s.rbegin(),s.rend()
#define RT(s) return cout<<s,0
#define watch(x) cout << (#x)<<" = "<<x<<endl
#define sz(s)    (int)(s.size())
#define PI acos(-1)
#define EPS 1e-100
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
int dy[] = { 1, -1, 0, 0, -1, 1, 1, -1 };
int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 };
void file() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
#else
	//freopen("street.in", "r", stdin);
	//freopen("out.txt", "w", stdout);
#endif
}
void fast() {
	std::ios_base::sync_with_stdio(0);
	cin.tie(NULL);
}
const int N = 1e5 + 9;
vi adjL[N];
int main() {
//	file();
	fast();
	int n, m; cin >> n >> m;
	vi hi(n);
	for (auto& it : hi)
		cin >> it;
	while (m--) {
		int u, v; cin >> u >> v;
		u--, v--;
		adjL[u].push_back(v);
		adjL[v].push_back(u);
	}
	int ans = 0;
	for (int i = 0; i < n; i++) {
		int mx = 0;
		for (auto& it : adjL[i])
			mx = max(mx, hi[it]);
		if (hi[i] > mx)
			ans++;
	}
	cout << ans << endl;
	return 0;
}