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>
using namespace std;
using ll = long long;
using ld = long double;
#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define FOR(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define REPD(i, n) for (ll i = (ll)(n)-1; i >= 0; i--)
const ll INF = 1LL << 60;
const int INF32 = 1 << 29;
const ll MOD = 1000000007;
using Graph = vector<vector<int>>; // グラフ型
using pint = pair<int, int>;

int main()
{
    int N, M;
    cin >> N >> M;
    vector<long long> H(N);
    REP(i, N)
    cin >> H[i];
    vector<int> G(N, 1);
    vector<bool> seen(N, false);
    REP(i, M)
    {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        if (H[a] > H[b])
        {
            G[b] = 0;
        }
        else if (H[a] == H[b])
        {
            G[a] = G[b] = 0;
        }
        else
        {
            G[a] = 0;
        }
    }

    ll ans = 0;
    REP(i, N)
    {
        ans += G[i];
    }
    cout << ans << endl;
}