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 <string.h>
#include <math.h>
#include <limits.h>

#define REP(i,n) for(i=0;i<n;i++)

#define MAX 1000000
#define MOD 1000000007

long long fac[MAX], finv[MAX], inv[MAX];

//prototype declaration
int i_in();
long long ll_in();
double d_in();
char c_in();
void s_in(char *s);
void i_out(int i);
void ll_out(long long i);
void d_out(double i);
void c_out(char c);
void s_out(char *s);
int gcd(int a, int b);
int lcm(int a, int b);
long long ll_gcd(long long a, long long b);
long long ll_lcm(long long a, long long b);
void sort_asc(int *a);
void sort_desc(int *a);
int asc(const void *a, const void *b);
int desc(const void *a, const void *b);
void ll_sort_asc(long long *a);
void ll_sort_desc(long long *a);
int ll_asc(const void *a, const void *b);
int ll_desc(const void *a, const void *b);
double to_deg(double rad);
double distance(int xi, int yi, int xj, int yj);
void initComb();
long long comb(int n, int k);

//main function
int main() {
    int n = i_in(), m = i_in(), i, a, b;
    int h[n], t[n];

    REP(i,n) {
        h[i] = i_in();
        t[i] = 1;
    }

    REP(i,m) {
        a = i_in();
        b = i_in();
        if (h[a-1] > h[b-1]) t[b-1] = 0;
        else if (h[a-1] < h[b-1]) t[a-1] = 0;
        else {
            t[a-1] = 0;
            t[b-1] = 0;
        }
    }

    a = 0;
    REP(i,n) if (t[i] == 1) a++;

    i_out(a);
    return 0;
}

//body function
int i_in() {
    int i;
    scanf("%d", &i);
    return i;
}

long long ll_in() {
    long long i;
    scanf("%lld", &i);
    return i;
}

double d_in() {
    double i;
    scanf("%lf", &i);
    return i;
}

char c_in() {
    char c;
    scanf("%c", &c);
    return c;
}

void s_in(char *s) {
    scanf("%s", s);
}

void i_out(int i) {
    printf("%d\n", i);
}

void ll_out(long long i) {
    printf("%lld\n", i);
}

void d_out(double i) {
    printf("%lf\n", i);
}

void c_out(char c) {
    printf("%c\n", c);
}

void s_out(char *s) {
    printf("%s\n", s);
}

/*
    int型の最大公約数
*/

int gcd(int a, int b) {
    int r;

    if (a < b) {
        r = a;
        a = b;
        b = r;
    }

    r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }

    return b;
}

/*
    int型の最小公倍数
*/

int lcm(int a, int b) {
    int r = ll_gcd(a, b);
    a /= r;
    return a * b;
}

/*
    long long型の最大公約数
*/

long long ll_gcd(long long a, long long b) {
    long long r;

    if (a < b) {
        r = a;
        a = b;
        b = r;
    }

    r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }

    return b;
}

/*
    long long型の最小公倍数
*/

long long ll_lcm(long long a, long long b) {
    long long r = ll_gcd(a, b);
    a /= r;
    return a * b;
}

/*
    int型の昇順ソート
*/

void sort_asc(int *a) {
    qsort(a, sizeof(a)/sizeof(int), sizeof(int), asc);
}

/*
    int型の降順ソート
*/

void sort_desc(int *a) {
    qsort(a, sizeof(a)/sizeof(int), sizeof(int), desc);
}

/*
    int型の昇順用比較関数
*/

int asc(const void *a, const void *b) {
    return *(int *)a - *(int *)b;
}

/*
    int型の降順用比較関数
*/

int desc(const void *a, const void *b) {
    return *(int *)b - *(int *)a;
}

/*
    long long型の昇順ソート
*/

void ll_sort_asc(long long *a) {
    qsort(a, sizeof(a)/sizeof(long long), sizeof(long long), ll_asc);
}

/*
    long long型の降順ソート
*/

void ll_sort_desc(long long *a) {
    qsort(a, sizeof(a)/sizeof(long long), sizeof(long long), ll_desc);
}

/*
    long long型の昇順用比較関数
*/

int ll_asc(const void *a, const void *b) {
    long long *A = (long long *)a;
    long long *B = (long long *)b;
    if (*A > *B) return 1;
    else if (*A < *B) return -1;
    else return 0;
}

/*
    long long型の降順用比較関数
*/

int ll_desc(const void *a, const void *b) {
    long long *A = (long long *)a;
    long long *B = (long long *)b;
    if (*A < *B) return 1;
    else if (*A > *B) return -1;
    else return 0;
}

/*
    弧度法を度数法に
*/

double to_deg(double rad) {
    return rad * 180.0 / (atan(1.0) * 4.0);
}

/*
    二点間の距離
*/

double distance(int xi, int yi, int xj, int yj) {
    return hypot(xj - xi, yj - yi);
}

/*
    組み合わせ用配列の初期化
*/

void initComb() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

/*
    nCkの組み合わせ計算
*/

long long comb(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}