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 <stdio.h>
#include <stdlib.h>

#include <iostream>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <stdint.h>
#include <string.h>

#define _USE_MATH_DEFINES
#include <math.h>

#include <vector>
#include <list>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <stack>
#include <deque>
#include <string>

#include <algorithm>
#include <functional>
#include <bitset>
#include <functional>
#include <chrono>
#include <random>

#define sqr(x) (x) * (x)

typedef unsigned int u32;
typedef int i32;
typedef unsigned long long int u64;
typedef long long int i64;
typedef uint16_t u16;
typedef int16_t i16;
typedef uint8_t u8;
typedef int8_t i8;

using namespace std;
using namespace std::chrono;

//const i64 mod = 1000000007ll;
const i64 smod = 998244353ll;
const i64 inf = 1'000'000'000'000'000'000ll + 7;
//const i64 inf = 1'000'000'007;

const long double eps = 1e-8;

const i64 MAXN = 100'000 + 5;

i64 T;

i64 n, k;

i64 a[MAXN];
vector<i64> e[MAXN];

int main(int argc, char* argv[]) {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cout.precision(15); cout.setf(ios::fixed);

    cin >> n >> k;
    for (i64 i = 1; i <= n; i++) {
        cin >> a[i];
    }

    for (i64 i = 0; i < k; i++) {
        i64 u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }

    i64 R = 0;
    for (i64 i = 1; i <= n; i++) {
        bool f = true;
        for (auto z : e[i]) {
            if (a[z] >= a[i]) {
                f = false;
                break;
            }
        }

        if (f) {
            R += 1;
        }
    }


    cout << R << endl;
    return 0;
}