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 <float.h>
 
#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define ll_cin(X) scanf("%lld", &X)
 
#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  ll_cout(X) printf("%lld\n",X)
 
typedef long long  ll;
#define rept(x, s, n) for (int x = s; x < n; x++)
 
#define S_RTN(S)  s_cout(S);return 0
 
 
#define _itoa(A,N) sprintf(A, "%d", N);
 
int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);
 
int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);
 
 
int s_asorts(const void* a, const void* b);
 
//昇順
int s_asorts(const void* a, const void* b) {
    return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
    return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
    return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
    int i;
    rept(i, 0, n) {
        l_cin2(*(a + i), *(b + i));
    }
    return 0;
}
int i_cins(int n, int* a) {
    int i;
    for (i = 0; i < n; i++) {
        i_cin(*(a + i));
    }
 
    return 0;
}
int _gcd(int a, int b) {
    int r, tmp;
    if (a < b) { tmp = a; a = b; b = tmp; }
    r = a % b;
    while (r) {
        if (!r) break;
        a = b; b = r;
        r = a % b;
    }
    return(b);
}
 
long _max(long a, long b) {
    return a > b ? a : b;
}
long _min(long a, long b) {
    return a < b ? a : b;
}
int _swp(int* a, int* b)
{
    int tmp;
    tmp = *b; *b = *a; *a = tmp;
    return 0;
}


int main(){
    int n,k;
    int cnt=0;
    scanf("%d%d",&n,&k);
    int a[n];
    int b[n];
    for(int i=0;i<n;i++){
        b[i]=0;
    }
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(int j=0;j<k;j++){
        int me1,me2,me3,me4;
        scanf("%d%d",&me3,&me4);
        
        me1=me3-1;
        me2=me4-1;
        if(a[me1]==a[me2]){
            b[me1]=1;
            b[me2]=1;
        }else if(a[me1]<a[me2]){
            b[me1]=1;
        }else{
            b[me2]=1;
        }
    }
        
    
    for(int i=0;i<n;i++){
        if(b[i]==0)
            cnt++;
    }
    printf("%d",cnt);
}