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 mod = 1000000007;
long long n, k, t, m, q, flag = 0;
long long po(long long k, long long n, long long ans, long long temp,
             long long ans1) {
  if (n == 0) return ans;
  while (temp <= n) {
    ans *= ans1;
    ans %= mod;
    ans1 = ans1 * ans1;
    ans1 %= mod;
    n = n - temp;
    temp *= 2;
  }
  return po(k, n, ans, 1, k) % mod;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    if (n <= 2) {
      cout << "YES"
           << "\n";
      continue;
    }
    long long temp;
    for (int i = 0; i < n; i++)
      if (a[i] == 1) temp = i;
    long long flag = 0;
    long long i = temp;
    if (a[(i + 1 + n) % n] - a[i] == 1) {
      long long f = 0;
      for (int j = 0; j < n - 1; j++) {
        if (a[(j + temp + 1 + n) % n] - a[(j + temp + n) % n] == 1)
          ;
        else {
          f = 1;
          break;
        }
      }
      if (f == 0)
        cout << "YES"
             << "\n";
      else
        cout << "NO"
             << "\n";
      continue;
    } else if (a[(i - 1 + n) % n] - a[i] == 1) {
      long long f = 0;
      for (int j = 0; j < n - 1; j++) {
        if (a[(-j + temp - 1 + n + n) % n] - a[(-j + temp + n + n) % n] == 1)
          ;
        else {
          f = 1;
          break;
        }
      }
      if (f == 0)
        cout << "YES"
             << "\n";
      else
        cout << "NO"
             << "\n";
      continue;
    } else
      cout << "NO"
           << "\n";
  }
  return 0;
}
