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)

#--Peaks
import itertools
from collections import Counter

N, M = map(int, input().split())
height = [int(i) for i in input().split()]

A = []
B = []
root = [[int(j) for j in input().split()] for i in range(M)]

judge = list(range(1,N+1)) # tennboudai no honsu

for i in range(M):
  q1 = root[i][0]
  q2 = root[i][1]
  if(height[q1-1] > height[q2-1]):
    judge[q2-1] = 0
  elif(height[q1-1] < height[q2-1]):
    judge[q1-1] = 0
  elif(height[q1-1] == height[q2-1]):
    judge[q1-1] = 0
    judge[q1-1] = 0

o = 2
c = Counter(tuple(x) for x in root)
double = [list(k) for k,v in c.items() if v==o]
number = len(double)

for i in range(number):
  q1 = double[i][0]
  q2 = double[i][1]
  judge[q1-1] = 0
  judge[q2-1] = 0

correct = [n for n in judge if n!=0]
#print(correct)
print(len(correct))