# 抄的
# 思路：随便找一个 2 的位置，循环往后移动n，确定每个位置与其后面位置大小的相关性，然后和 a 进行检查
import sys, time, random
from collections import deque, Counter, defaultdict
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
inf = 2 ** 63 - 1
mod = 998244353

def solve():
    n = ii()
    a = li()
    if 2 not in a:
        return 'No'
    p = a.index(2)
    
    d = [None] * n
    for i in range(p, p + n):
        i %= n
        to = None
        if a[i] == 2:
            to = '>'
        if a[i] == 0:
            to = '<'
        if a[i] == 1:
            to = d[(i - 1) % n]
        d[i] = to
    for i in range(n):
        if a[i] != (d[(i - 1) % n] == '<') + (d[i % n] == '>'):
            return 'No'
    return 'Yes'

for _ in range(ii()):
    print(solve())