# ACだがa条件で有向辺を張る、という前回の方針でやってみる
# 有向辺にすれば閉路はできない、Ai<Ajだから
# すると入次数が0のところからスタートすればいい

N, M = map(int, input().split())
A = list(map(int, input().split()))
edges = [[] for i in range(N)]
inward = [0]*N
for i in range(M):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    if A[u] < A[v]:
        edges[u].append(v)
        inward[v] += 1
    elif A[v] < A[u]:
        edges[v].append(u)
        inward[u] += 1

K = int(input())
B = list(map(int, input().split()))
switch = [0]*N
for b in B:
    switch[b-1] = 1

from collections import deque
que = deque()
for i in range(N):
    if inward[i] == 0:
        que.append(i)

ans_list = []
while que:
    current = que.popleft()
    if switch[current] == 1:
        ans_list.append(current)
        for nxt in edges[current]:
            switch[nxt] ^= 1
            inward[nxt] -= 1
            if inward[nxt] == 0:
                que.append(nxt)
    elif switch[current] == 0:
        for nxt in edges[current]:
            inward[nxt] -= 1
            if inward[nxt] == 0:
                que.append(nxt)        

print(len(ans_list))
for a in ans_list:
    print(a+1)