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.Arrays;
import java.util.Scanner;

public class TaskA {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int testCases = scanner.nextInt();
        for (int i = 0; i < testCases; i++) {
            int n = scanner.nextInt();
            int[] arr = new int[n], ans = new int[n];
            for (int j = 0; j < n; j++) {
                arr[j] = scanner.nextInt();
            }
            int idx = 0;
            for (int j = 0; j < n; j++) {
                if (arr[j] == 1) {
                    idx = j; break;
                }
            }
            boolean isLeftSorted = isSorted(0, idx - 1, arr);
            boolean isRightSorted = isSorted(idx + 1, n - 1, arr);
            if (isLeftSorted && isRightSorted)
                System.out.println("YES");
            else System.out.println("NO");
        }
    }

    private static boolean isSorted(int left, int right, int[] arr) {
        if (left == right) return true;
        if (left > right) return true;
        boolean sorted = true;
        if (arr[left] < arr[right]) {
            for (int i = left; i < right; i++) {
                if (arr[i] > arr[i +1]) {
                    sorted = false; break;
                }
            }
        } else {
            for (int i = left; i < right; i++) {
                if (arr[i] < arr[i +1]) {
                    sorted = false; break;
                }
            }
        }
        return sorted;
    }
}