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: Wrong Answer)

N, M = map(int, input().split())
H_list = list( map(int, input().split()) )

AB_list = list()

for _ in range(M):
    A,B = map(int, input().split())
    AB_list.append( (A, B) )

#print(AB_list)
bad_towers = dict()
can_good_towers = dict()

for (A,B) in AB_list:
    win = A
    lose = B
    if H_list[A-1] < H_list[B-1]:
        win = B
        lose = A
    
    # try:
    #     if can_good_towers[lose] == 1:
    #         del(can_good_towers[lose])
    # except KeyError:
    #     pass
        
    can_good_towers[str(win)] = 1
    bad_towers[str(lose)] = 1

#print(can_good_towers)
#print(bad_towers)
# 一度もルートの組み合わせに入らず、またbadに入らないものの個数が良いやつ
good_tower_num = N - len(bad_towers)
print(good_tower_num)