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;
vector<long long> adj[1000000];
long long A[10000], n, i, x, y, k, j, flag, t;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    flag = 0;
    cin >> n;
    long long A[n + 2];
    memset(A, 0, sizeof(A));
    for (i = 1; i <= n; i++) {
      cin >> A[i];
      if (A[i] == 1) x = i;
    }
    if (x == n)
      y = 1;
    else
      y = x + 1;
    if (x == 1)
      k = n;
    else
      k = x - 1;
    if (A[y] == 2) {
      j = 1;
      for (i = x; i <= n; i++) {
        if (A[i] == (j++))
          continue;
        else {
          flag = 1;
          break;
        }
      }
      for (i = 1; i < x; i++) {
        if (A[i] == (j++))
          continue;
        else {
          flag = 1;
          break;
        }
      }
    } else if (A[k] == 2) {
      j = 1;
      for (i = x; i >= 1; i--) {
        if (A[i] == (j++))
          continue;
        else {
          flag = 1;
          break;
        }
      }
      for (i = n; i > x; i--) {
        if (A[i] == (j++))
          continue;
        else {
          flag = 1;
          break;
        }
      }
    } else {
      flag = 1;
    }
    if (flag)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
    }
  }
}
