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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define lli long long
#define mp make_pair
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define allr(v) (v).rbegin(), (v).rend()
#define ok return 0;
#define MOAE                 \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
map<int, int> vis;
set<ll> s;
int main()
{
    MOAE;
    ll n, m, ans = 0;
    cin >> n >> m;
    vector<ll> h(n);
    map<ll, vector<ll>> st;
    for (auto &i : h)
        cin >> i;
    while (m--)
    {
        ll a, b;
        cin >> a >> b;
        a--, b--;
        st[a].push_back(b);
        st[b].push_back(a);
    }
    for (auto it : st)
    {
        bool f = true;
        for (auto i : it.second)
        {
            if (h[it.first] < h[i])
            {
                f = false;
                break;
            }
        }
        if (f)
            ans++;
    }
    cout << ans << endl;
}