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,cc,n) for(int i=cc;i<(n);++i)
#define drep(i,cc,n) for(int i=cc;i>(n);--i)
using namespace std;
using ll = long long;

int main()
{
	int N, M, count = 0,max=0;
	//string S;
	cin >> N >> M;
	vector<ll> H(N);
	vector<int> A(M), B(M), c(N);//c = hens
	vector<vector<int>> hen(N, vector<int>(N));
	rep(i, 0, N)
	{
		cin >> H[i];
	}
	rep(i, 0, M)
	{
		cin >> A[i] >> B[i];
		++c[A[i] - 1];
		++c[B[i] - 1];
		hen[A[i]][c[A[i]]] = B[i];
	}
	
	rep(i, 0, N)
	{
		if (c[i] == 0)
			++count;
		else
		{
			max = 1;
			rep(j, 0, c[i])
			{
				if (H[i] <= H[hen[i][j]])
					max = 0;
			}
			if (max > 0)
				count++;
		}
	}
	cout << count << endl;
}