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
# -*- coding: utf-8 -*-
"""
Created on Wed May 20 16:14:12 2020

@author: dell
"""

def round(l):
    t=len(l)
    if t==1:
        return 1
    
    i=l.index(1)
    j=l.index(2)        
    #print(i,j)
    if i-j==-1 or i-j== t-1:
        p=j
        q=(p+1)%t
        
        while l[p]-l[q]==-1:
             p=(p+1)%t
             q=(p+1)%t
             
             if q==i:
                 
                 return 1
        else:
            return 0
    
    elif i-j==1:
        p=j
        q=p-1
        
        while l[p]-l[q]==-1:
            
            p=p-1
            q=p-1
            if q==-(t-i):
                return 1
        else:
            return 0
        
    elif i-j== 1-t:
        p=-1
        q=p-1
        
        while l[p]-l[q]==-1:
            p=p-1
            q=p-1
            if q==-(t-i):
                return 1
        else:
            return 0
    


q=int(input())
m=[]
for i in range(q):
    n=int(input())
    l=[x for x in map(int,input().split())]
    if len(l)==n:
        k=round(l)
        m.append(k)
for i in m:
        if i==1:
            print('YES')
        else:
            print('NO')