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

import sys
import math
import time
#import numpy as np
import collections
from collections import deque
from collections import Counter
import queue
import copy
import bisect
import heapq
import itertools


#sys.setrecursionlimit(10**7)
#N, Q = map(int, input().split())
#G = [list(input()) for i in range(H)]
#INF = V * 10001
#A = [int(i) for i in input().split()]
#AB = [list(map(int, input().split())) for _ in range(K)]


N, M = map(int, input().split())
H = [int(i) for i in input().split()]
AB = [list(map(int, input().split())) for i in range(M)]
s={AB[0][0]}
A = [0]*N
ans = 0
#print(H)
for i in range(M):
    s.add(AB[i][0])
    s.add(AB[i][1])
    #print(AB[i][0]-1, AB[i][1]-1)
    if(H[AB[i][0]-1]==H[AB[i][1]-1]):
        A[AB[i][0]-1] =0
        A[AB[i][1]-1] =0
    elif(H[AB[i][0]-1]<H[AB[i][1]-1]):
        A[AB[i][0]-1] =0
        A[AB[i][1]-1] =1
    else:
        A[AB[i][0]-1] =1
        A[AB[i][1]-1] =0

ans = A.count(1)
ans += N-len(s)


print(ans)