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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

## coding: UTF-8
N,M = map(int,input().split())
H = list(map(int,input().split()))

path = []
for i in range(M):
    path.append(list(map(int,input().split())))

l = [] 
for i in range(N):
    l.append([0] * N + [H[i]])



highest = [0] + [1] * N
#print(highest)
#print(H)
for i in range(M):
    #print('path[i][0] = {}, path[i][1] = {}'.format(path[i][0], path[i][1]))
    #print('H[path[i][0]-1]:{}, H[path[i][1]-1]:{}'.format(H[path[i][0]-1], H[path[i][1]-1]))
    p = path[i][1]-1
    q = path[i][0]-1
    x = H[p]
    y = H[q]
    #if( x >= y ):
        #print('deny at {}'.format(path[i][0]))
        #highest[q+1] = 0
    #if(  x <= y  ):
        #print('deny at {}'.format(path[i][1]))
        #highest[p+1] = 0
    #print('highest:{}'.format(highest))

print(sum(highest))

'''
for i in range(M):
    print('path[i][0] = Ai = {}, path[i][1] = Bi = {}'.format(path[i][0], path[i][1]))
    l[ path[i][0]-1 ][ path[i][1]-1 ] = H[path[i][1]-1]
    l[ path[i][1]-1 ][ path[i][0]-1 ] = H[path[i][0]-1]
    print(l)
'''



'''
m_neigh = []
for i in range(N):
    print(l[i][:N])
    m_neigh.append(  max(l[i][:N] )  )
    print(m_neigh)
'''
'''
cnt = 0
for i in range(N):
    
'''