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)

from sys import stdin
input = stdin.readline

#二次元配列から重複を取り除く関数
def get_unique_list(seq) :
    seen = []
    return [x for x in seq if x not in seen and not seen.append(x)]

n, m = map(int, input().split()) # 展望台と道の数
h_list = list(map(int, input().split())) # n個の展望台の標高
# どの展望台とどの展望台を結ぶ道があるかのリスト
plane_load_list = []
for i in range(m) :
    plane_load_list.append(list(map(int, input().split())))

# 全く同じ道(向きもいっしょ)は2回以上判定する必要がないので取り除く
load_list = get_unique_list(plane_load_list)
# 

num = [x for x in range(1, n+1)]
bad = []
for i in range(0, len(load_list)) :
    if load_list[i][0] not in bad or load_list[i][1] not in bad :
        if h_list[load_list[i][0]-1]<h_list[load_list[i][1]-1] :
            bad.append(load_list[i][0])
        elif h_list[load_list[i][0]-1]>h_list[load_list[i][1]-1] :
            bad.append(load_list[i][1])
        else :
            bad.append(load_list[i][0])
            bad.append(load_list[i][1])

bad_value = list(set(bad))
print(len(num)-len(bad_value))