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.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Scanner;

public class Main {
	static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));
	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
//		long tic = System.currentTimeMillis();
//		long toc = System.currentTimeMillis();

		int t = nextInt();
		while (t-- > 0) {
			boolean flag = true;
			int n = nextInt();
			int arr[] = new int[n];
			int id = 0;
			for (int i = 0; i < n; i++) {
				arr[i] = nextInt();
				if (arr[i] == 1) {
					id = i;
				}
			}
			if (arr[(id - 1 + n) % n] == arr[id] + 1) {
				for (int i = 0; i < n; i++) {
					if (arr[(id - i + n) % n] != arr[id] + i) {
						flag = false;
						break;
					}
				}
			} else if (arr[(id + 1) % n] == arr[id] + 1) {
				for (int i = 0; i < n; i++) {
					if (arr[(id + i) % n] != arr[id] + i) {
						flag = false;
					}
				}
			}
			if (flag)
				System.out.println("YES");
			else
				System.out.println("NO");
		}

//		System.out.println("Elapsed time: " + (toc - tic) + " ms");
	}

	static long pow(long a, long b, long mod) {
		if (b == 0)
			return 1;
		if (b == 1)
			return a;
		if ((b & 1) == 0)
			return pow(a * a % mod, b >> 1, mod);
		else
			return a * pow(a * a % mod, b >> 1, mod) % mod;
	}

	static int nextInt() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}

	static double nextDouble() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.nval;
	}

	static String next() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.sval;
	}

	static long nextLong() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (long) st.nval;
	}
}