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.Scanner;

public class CF1203A {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		for (int i = 0; i < n; i++) {
			int m = s.nextInt();
			int[] v = new int[m];
			int initial = -1;
			for (int j = 0; j < v.length; j++) {
				int x =s.nextInt();
				v[j]=x;
				if(x==1) {
					initial=j;
				}
			}
			if(m==1) {
				System.out.println("YES");
			}else {
				String ans = "YES";
				ans=valideClockWise(v,initial);
				if(ans.equals("NO")) {
					ans = valideCounterClockWise(v, initial);
				}
				System.out.println(ans);
			}
		}

	}
	public static String valideClockWise(int[]v,int initial) {
		String ans = "YES";
		boolean complete = true;
		if(initial<v.length-1) {
			for (int i = initial+1; i < v.length; i++) {
				if(v[i]!=(v[i-1]+1)) {
					ans="NO";
					complete=false;
					break;
				}
			}
		}
		if(complete) {
			for (int i = 1; i < initial; i++) {
				if(v[i]!=(v[i-1]+1)) {
					ans="NO";
					break;
				}
			}
		}
		return ans;
	}
	public static String valideCounterClockWise(int[]v,int initial) {
		String ans = "YES";
		int dif= v.length-initial;
		boolean complete = true;
		if(initial>0) {
			for (int i = initial-1; i >=0; i--) {
				if(v[i]!=(v[i+1]+1)) {
					ans="NO";
					complete=false;
					break;
				}
			}
		}
		if(complete) {
			for (int i = v.length-2; i >initial; i--) {
				if(v[i]!=(v[i+1]+1)) {
					ans="NO";
					break;
				}
			}
		}
		return ans;
	}
}
