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
import java.util.HashSet;
import java.util.Scanner;

public class Solution {

    private static boolean isClock(int[] a, int onepos) {
        int prev = onepos;
        int next = (prev==a.length-1) ? 0 : onepos+1;
        while (next != onepos) {
            if (a[prev] + 1 == a[next]) {
                prev = next;
                ++next;
            }
            else {
                //System.out.println(prev + " " + next);
                return false;
            }
            if (next == a.length) next = 0;
        }
        return true;
    }

    private static boolean isCounterClock(int[] a, int npos) {
        int prev = npos;
        int next = npos+1;
        while (next != npos) {
            if (next == a.length) next = 0;
            if (a[prev] - 1 == a[next]) {
                prev = next;
                ++next;
            }
            else return false;
        }
        return true;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int q = sc.nextInt();
        while (q-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nextInt();
            int onepos = -1, npos = -1;
            for (int i = 0; i < a.length; i++) {
                if (a[i] == 1) {
                    onepos = i;
                }
                if (a[i] == n) npos = i;
            }
            //System.out.println(npos!=-1);
            if (isClock(a, onepos) || isCounterClock(a, npos))
                System.out.println("YES");
            else System.out.println("NO");
        }
    }
}