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: Runtime Error)

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

#define A 100

long  reach[A][A];

int main(void){
    int numdesk,numroad;

    scanf("%d %d",&numdesk,&numroad);

    int *height,**road;

    height=(int *)malloc(sizeof(int)*numdesk);

    for(int i=0;i<numdesk;i++) scanf("%d",height[i]);

    road=(int **)malloc(sizeof(int *)*numroad);
    for(int i=0;i<numroad;i++) road[i]=(int *)malloc(sizeof(int)*2);


    for(int i=0;i<A;i++){
        for(int j=0;j<A;j++) reach[i][j]=0;
    }
    int num[A]={0};

    for(int i=0;i<numroad;i++){
        scanf("%ld %ld",road[i][0],&road[i][1]);
        reach[road[i][0]][num[road[i][0]]]=road[i][1];
        reach[road[i][1]][num[road[i][1]]]=road[i][0];
    }

    int maxID[A]={0};
    int maxH=0;
    int flag[A]={0};

    for(int i=0;i<numdesk;i++){
        int k=0;
        maxID[i]=i+1;
        maxH=height[i];
        while(reach[i][k]!=0){
            if(height[reach[i][k]]>maxH) {
                maxH=height[reach[i][k]];
                maxID[i]=reach[i][k];
            }
            k++;
        }
        flag[maxID[i]]=1;
    }

    int count=0;
    
    for(int i=0;i<numdesk;i++){
        count+=flag[i];
    }

    printf("%d\n",count);


    return 0;
}