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: Wrong Answer)

#include <stdio.h>

struct yama {
  int height;   //この山の高さ
  int highest;  //
};

int main(void) {
  int i, n, m, h;  //展望台の数n,道の数m
  scanf("%d %d", &n, &m);
  struct yama yam[100000];
  for (i = 0; i < n; i++) {
    scanf("%d", &h);
    yam[i].height = h;    //こいつの高さ.高い方に更新
    yam[i].highest = -1;  //他と接続なしで-1(いい展望台確定)
  }
  int a, b, atmp, btmp;
  for (i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    a = a - 1;  //補正
    b = b - 1;
    atmp = a;
    btmp = b;
    while (yam[atmp].highest != -1) {
      atmp = yam[atmp].highest;
    }
    while (yam[btmp].highest != -1) {
      btmp = yam[btmp].highest;
    }  // a,bをそれぞれの近くで一番高い山にする
    if (yam[atmp].height > yam[btmp].height) {
      yam[b].highest = atmp;
    } else {
      yam[a].highest = btmp;
    }
  }
  int count = 0;
  for (i = 0; i < n; i++) {
    // printf("height : %d\nhighest : %d\n", yam[i].height, yam[i].highest);
    if (yam[i].highest == -1) {
      count++;
    }
  }
  printf("%d\n", count);
  return 0;
}