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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

import copy
n, m = map(int, input().split()) #n: 展望台の数, m: 道の数
height = list(map(int, input().split())) #高さリスト。nコある。
path = [list(map(int, input().split())) for _ in range(m)] #道リスト。
good_tower = 0 #解答

for i in range(n): #各展望台について.i+1の周辺をサーチ。
  h_i = height[i]
  con_num = [] #隣接する台番号を格納
  for j in range(m): #全pathをサーチ
    if i+1 in path[j]:
      p = copy.deepcopy(path[j])
      p.remove(i+1)
      c = p[0]
      con_num.append(c)
  if len(con_num) == 0:
    good_tower += 1
  else:
    con_num = list(set(con_num))
    cl = len(con_num)
    h_near_l = []
    for k in range(cl):
      t_num = con_num[k] #隣接台のナンバー
      h_near = height[t_num-1]
      h_near_l.append(h_near)
    if all(h_i > h_near_l[i] for i in range(cl)):
      good_tower += 1
      
print(good_tower)