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
class linked:
    def __init__(self,a):
        self.a=a
def right_trav(head,n):
    temp=head
    for i in range(n-1):
        # print('curr=',temp.a,'right=',temp.right.a)
        if (temp.a)+1!=(temp.right.a):
            break
        temp=temp.right
    # print('i=',i)
    if i==n-2:
        return True
    else:
        return False

def left_trav(head,n):
    temp=head
    for i in range(n-1):
        # print('curr=',temp.a,'left=',temp.left.a)
        if (temp.a)+1!=(temp.left.a):
            break
        temp=temp.left
    if i==n-2:
        return True
    else:
        return False

for t in range(int(input())):
	n=int(input())
	a=[int(x) for x in input().split()]
	if n!=1:
		head=linked(a[0])
		temp=head
		for i in range(1,n):
		    temp.right=linked(a[i])
		    temp.right.left=temp
		    temp=temp.right
		temp.right=head
		head.left=temp
		temp=head
		c=1
		while temp.a!=1:
		    temp=temp.right
		    c+=1
		# print('at c =',c,'temp.date=',temp.a)
		if(left_trav(temp,n) or right_trav(temp,n)):
			print('YES')
		else:
			print('NO')
	else:
		print('YES')