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>


typedef struct
{
    int height;
    int targetMaxHeight;
}Info;

#define DEBUG


int main()
{
    int i, j, k, m;

    char str[1000];
    fgets(str, sizeof(str), stdin);

    char* temp;
    temp = strtok(str, " \n");
    // 展望台の数
    int TenbodaiNum = atoi(temp);
    Info* pInfo = (Info*)malloc(sizeof(Info) * TenbodaiNum );
    memset(pInfo, 0, sizeof(Info) * TenbodaiNum);


    // 道の数
    temp = strtok(NULL, " \n");
    int PathNum = atoi(temp);

    //高さの取得

    fgets(str, sizeof(str), stdin);

    int hoge = 0;
    temp = strtok(str, " \n");

    while (temp != NULL)
    {
        pInfo[hoge].height = atoi(temp);
        hoge++;
        temp = strtok(NULL, " \n");
    }

    //道情報取得
    for (i = 0;i < PathNum;i++)
    {
        int iPos1;
        int iPos2;
        fgets(str, sizeof(str), stdin);
        temp = strtok(str, " \n");
        iPos1 = atoi(temp) -1;

        temp = strtok(NULL, " \n");
        iPos2 = atoi(temp) -1;

        if (pInfo[iPos1].targetMaxHeight < pInfo[iPos2].height)
        {
            pInfo[iPos1].targetMaxHeight = pInfo[iPos2].height;
        }
        if (pInfo[iPos2].targetMaxHeight < pInfo[iPos1].height)
        {
            pInfo[iPos2].targetMaxHeight = pInfo[iPos1].height;
        }
    }

    // 判定
   
    int iGoodNum = 0;

    for (i = 0;i < TenbodaiNum;i++)
    {
        if (pInfo[i].height > pInfo[i].targetMaxHeight) iGoodNum++;
    }



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


    return 0;
}