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: Accepted)

#import sys
MOD = 10 ** 9 + 7
INFI = 10**10
#input = sys.stdin.readline
import math

import itertools
import heapq
#import bisect

import copy
from functools import lru_cache

def sosuhante(n):
    for k in range(2, int(math.sqrt(n))+1):
        if n% k ==0:
            return False
    return True
def cmb(n, r):
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n

    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]

    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot

    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])

    return result

def kingaku(a,b,n):
    keta=len(str(n))
    return a*n+b*keta

def my_index(l, x, default=False):
    if x in l:
        return l.index(x)
    else:
        return default

#    h,w,a,b = map(int, input().split())
#    c = [[0 for j in range(n)] for i in range(n)]

def ret(a):
    c=[None]*(len(a)-1)
    if len(a)==1:
        return a[0]
    elif len(a)==0:
        return 0
    for i in range(1,len(a)):
        c[i-1]=abs(a[i]-a[i-1])
    return ret(c)


def main():
    n,m=map(int,input().split())
    h=list(map(int,input().split()))
    a=[[h[i]] for i in range(n)]
    for i in range(m):
        x,y=map(int,input().split())
        a[x-1].append(h[y-1])
        a[y-1].append(h[x-1])
    total=0
    for i in range(n):
        if len(a[i])==1:
            total+=1
        elif max(a[i][1:])<a[i][0]:
            total+=1
    print(total)






if __name__ == "__main__":

    main()