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

#define mod 1000000007
#define limit 200001
typedef long long ll;
typedef long double ld;
#define rep(i,n) for(i = 0; i < n; i++)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
int up_int(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_int(const void *a, const void *b){return *(int*)b - *(int*)a;}
int up_char(const void *left, const void *right) {return strcmp((char *)left,(char *)right);}

int main(){
    int n,m,i,j,count=0;
    int h[100001];
    int a[100001];
    int b[100001];
    int flag[100001] = {0};
    
    scanf("%d %d",&n,&m);
    for(i = 1; i <= n; i++) scanf("%d",&h[i]);
    for(i = 1; i <= m; i++) scanf("%d %d",&a[i],&b[i]);
    
    for(i = 1; i <= m; i++){
        if(h[a[i]] > h[b[i]]){
            flag[b[i]]--;
//            printf("%d is bad\n",b[i]);
        }else if(h[a[i]] == h[b[i]]){
            flag[a[i]]--;
            flag[b[i]]--;
//            printf("%d and %d are bad\n",a[i],b[i]);
        }else{
            flag[a[i]]--;
//            printf("%d is bad\n",a[i]);
        }
    }
    for(i = 1; i <= n; i++){
        if(flag[i] == 0){
            count++;
//            printf("%d is good tower!\n",i);
        }
    }
//    for(i = 1; i <= n; i++) printf("%d ",flag[i]);
    printf("%d\n",count);
    return 0;
}