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)

#MLE注意！0や1のケースに注意

def main():
    import sys
    input = sys.stdin.readline
    sys.setrecursionlimit(10**7)
    from collections import Counter, deque
    #from collections import defaultdict
    from itertools import combinations, permutations, accumulate, groupby, product
    from bisect import bisect_left,bisect_right
    from heapq import heapify, heappop, heappush
    import math
    #from math import gcd

    #inf = 10**17
    #mod = 10**9 + 7

    n,m = map(int, input().split())
    h = list(map(int, input().split()))
    adj = [[] for _ in range(n)] #頂点数, 場合によって変える
    for _ in range(m):
        a,b = map(int, input().split())
        adj[a-1].append(b-1)
        adj[b-1].append(a-1)
    #-1まだ　0負け確　1良い
    check = [-1]*n
    for i in range(n):
        if check[i] == 0:
            continue
        if len(adj[i]) == 0:
            check[i] = 1
        else:
            for j in adj[i]:
                if h[i] < h[j]:
                    check[i] = 0
                    break
                elif h[i] == h[j]:
                    check[i] = 0
                    check[j] = 0
                    break
                else:
                    check[j] = 0
            if check[i] == -1:
                check[i] = 1
    res = 0
    for i in check:
        if i == 1:
            res += 1
    print(res)


if __name__ == '__main__':
    main()