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
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = res * x;
    x = x * x;
    n = n / 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, i;
    cin >> n;
    int a[n], x, y;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] == n) x = i;
      if (a[i] == 1) y = i;
    }
    int flag = 1;
    if (x == 0 && y == n - 1) {
      for (i = 1; i < n; i++) {
        if (a[i] + 1 != a[i - 1]) {
          flag = 0;
          break;
        }
      }
      if (flag == 0)
        cout << "NO\n";
      else
        cout << "YES\n";
    } else {
      if (x == n - 1 && y == 0) {
        for (i = 0; i < n - 1; i++) {
          if (a[i] + 1 != a[i + 1]) {
            flag = 0;
            break;
          }
        }
        if (flag == 0)
          cout << "NO\n";
        else
          cout << "YES\n";
      } else {
        if (abs(x - y) == 1) {
          if (x > y) {
            int flag1 = 1, flag2 = 1;
            for (i = x + 1; i < n; i++) {
              if (a[i] + 1 != a[i - 1]) {
                flag1 = 0;
                break;
              }
            }
            for (i = y - 1; i >= 0; i--) {
              if (a[i] != a[i + 1] + 1) {
                flag2 = 0;
                break;
              }
            }
            if (flag1 == 1 && flag2 == 1)
              cout << "YES\n";
            else
              cout << "NO\n";
          } else {
            int flag1 = 1, flag2 = 1;
            for (i = y + 1; i < n; i++) {
              if (a[i] != a[i - 1] + 1) {
                flag1 = 0;
                break;
              }
            }
            for (i = x - 1; i >= 0; i--) {
              if (a[i] + 1 != a[i + 1]) {
                flag2 = 0;
                break;
              }
            }
            if (flag1 == 1 && flag2 == 1)
              cout << "YES\n";
            else
              cout << "NO\n";
          }
        } else {
          cout << "NO\n";
        }
      }
    }
  }
}
