There are n students standing in a circle in some order. The index of the i-th student is p_i. It is guaranteed that all indices of students are distinct integers from 1 to n (i. e. they form a permutation).

Students want to start a round dance. A clockwise round dance can be started if the student 2 comes right after the student 1 in clockwise order (there are no students between them), the student 3 comes right after the student 2 in clockwise order, and so on, and the student n comes right after the student n - 1 in clockwise order. A counterclockwise round dance is almost the same thing — the only difference is that the student i should be right after the student i - 1 in counterclockwise order (this condition should be met for every i from 2 to n). 

For example, if the indices of students listed in clockwise order are [2, 3, 4, 5, 1], then they can start a clockwise round dance. If the students have indices [3, 2, 1, 4] in clockwise order, then they can start a counterclockwise round dance.

Your task is to determine whether it is possible to start a round dance. Note that the students cannot change their positions before starting the dance; they cannot swap or leave the circle, and no other student can enter the circle. 

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.

The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of students.

The second line of the query contains a permutation of indices p_1, p_2, ..., p_n (1 ≤ p_i ≤ n), where p_i is the index of the i-th student (in clockwise order). It is guaranteed that all p_i are distinct integers from 1 to n (i. e. they form a permutation).

Output

For each query, print the answer on it. If a round dance can be started with the given order of students, print "YES". Otherwise print "NO".

Example

Input


5
4
1 2 3 4
3
1 3 2
5
1 2 3 5 4
1
1
5
3 2 1 5 4


Output


YES
YES
NO
YES
YES
#from statistics import median
#import collections
#aa = collections.Counter(a) # list to list || .most_common(2)で最大の2個とりだせるお a[0][0]
from fractions import gcd
from itertools import combinations,permutations,accumulate, product # (string,3) 3回
#from collections import deque
from collections import deque,defaultdict,Counter
import decimal
import re
import math

#
#
#
# pythonで無理なときは、pypyでやると正解するかも！！
#
#
# my_round_int = lambda x:np.round((x*2 + 1)//2)
# 四捨五入g
#
# インデックス系
# int min_y = max(0, i - 2), max_y = min(h - 1, i + 2);
# int min_x = max(0, j - 2), max_x = min(w - 1, j + 2);
#
#
import sys
sys.setrecursionlimit(10000000)
mod = 10**9 + 7
#mod = 9982443453
#mod = 998244353
from sys import stdin
readline = stdin.readline
def readInts():
  return list(map(int,readline().split()))
def readTuples():
    return tuple(map(int,readline().split()))
def I():
  return int(readline())
for _ in range(I()):
    n = I()
    A = readInts()
    if n == 1:
        print('YES')
        continue
    # 1 を探せ
    idx = None
    for i in range(n):
        if A[i] == 1:
            idx = i
            break
    left = False
    right = False
    if A[(idx+1)%n] == 2:
        right = True
    elif A[idx-1] == 2:
        left = True

    # 左方向に
    new = None
    if left:
        new = A[idx+1:] + A[:idx+1]
    elif right:
        new = A[idx:] + A[:idx]
    rela = list(range(1,n+1)) if right else list(range(n,0,-1))
    print('YES' if new == rela else 'NO')
