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 connect;
    int* connctTarget;
}Info;



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);
    for (i = 0; i < TenbodaiNum;i++)
    {
        pInfo[i].connect = 0;
        pInfo[i].connctTarget = NULL;
    }



    // 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);

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

        int same = 0;
        for (j = 0;j < pInfo[iPos1 - 1].connect;j++)
        {
            if (pInfo[iPos1 - 1].connctTarget[j] == iPos2 - 1)
            {
                same = 1;
                break;
            }

        }
        if (same == 1) continue;

        pInfo[iPos1 - 1].connect++;
        pInfo[iPos1 - 1].connctTarget = (int*)realloc(pInfo[iPos1 - 1].connctTarget, sizeof(int) * pInfo[iPos1 - 1].connect);
        pInfo[iPos1 - 1].connctTarget[pInfo[iPos1 - 1].connect - 1] = iPos2 - 1;

        pInfo[iPos2 - 1].connect++;
        pInfo[iPos2 - 1].connctTarget = (int*)realloc(pInfo[iPos2 - 1].connctTarget, sizeof(int) * pInfo[iPos2 - 1].connect);
        pInfo[iPos2 - 1].connctTarget[pInfo[iPos2 - 1].connect - 1] = iPos1 - 1;

    }

    // 判定
    int iGood = 1;   //1:Good
    int iGoodNum = 0;

    for (i = 0;i < TenbodaiNum;i++)
    {
        int selfHeight = pInfo[i].height;
        for (j = 0;j < pInfo[i].connect;j++)
        {
            if (selfHeight <= pInfo[pInfo[i].connctTarget[j] - 1].height)
            {
                iGood = 0;
                break;
            }
        }

        if (iGood == 1) iGoodNum++;
        iGood = 1;
    }



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


    return 0;
}