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;
int main() {
  long long a, b, c, d, e, f, n;
  long long v[1000];
  cin >> a;
  for (int i = 1; i <= a; i++) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> b;
      if (i == 1) {
        v[n + 1] = b;
      }
      v[i] = b;
    }
    v[0] = b;
    if (n == 1) {
      cout << "YES" << endl;
      continue;
    }
    int sw = 0;
    if (((v[1] == n and v[2] == 1) or v[1] < v[2]) and
        (v[n - 1] < v[n] or (v[n - 1] == n and v[n] == 1))) {
      for (int i = 1; i <= n; i++) {
        if (v[i] == 1) {
          if (v[i - 1] != n or v[i + 1] != v[i] + 1) {
            sw++;
            break;
          }
          continue;
        }
        if (v[i] == n) {
          if (v[i - 1] != v[i] - 1 or v[i + 1] != 1) {
            sw++;
            break;
          }
          continue;
        }
        if (v[i - 1] != v[i] - 1 or v[i + 1] != v[i] + 1) {
          sw++;
          break;
        }
      }
      if (sw == 0) {
        cout << "YES" << endl;
      } else
        cout << "NO" << endl;
      continue;
    }
    if (((v[1] == 1 and v[2] == n) or v[1] > v[2]) and (v[n - 1] > v[n]) or
        (v[n - 1] == 1 and v[n] == n)) {
      for (int i = 1; i <= n; i++) {
        if (v[i] == 1) {
          if (v[i - 1] != v[i] + 1 or v[i + 1] != n) {
            sw++;
            break;
          }
          continue;
        }
        if (v[i] == n) {
          if (v[i - 1] != 1 or v[i + 1] != v[i] - 1) {
            sw++;
            break;
          }
          continue;
        }
        if (v[i - 1] != v[i] + 1 or v[i + 1] != v[i] - 1) {
          sw++;
          break;
        }
      }
      if (sw == 0) {
        cout << "YES" << endl;
      } else
        cout << "NO" << endl;
      continue;
    }
    cout << "NO" << endl;
  }
}
