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;
map<long long, long long> freq;
void dom(long long x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      freq[i]++;
      while (x % i == 0) x /= i;
    }
  if (x > 1) freq[x]++;
}
bool chech(long long r, long long g, long long b, long long w) {
  long long gg = r % 2 + g % 2 + b % 2 + w % 2;
  if (gg < 2 && (r >= 0 && g >= 0 && b >= 0))
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, r, g, b, w;
  cin >> n;
  while (n--) {
    cin >> r;
    long long x[204] = {};
    for (int i = 0; i < r; ++i) cin >> x[i];
    bool t = true;
    for (int i = 0; i < r - 1; ++i) {
      if (abs(x[i] - x[i + 1]) == 1)
        continue;
      else {
        t = false;
        break;
      }
    }
    if (t == true)
      cout << "YES\n";
    else {
      if (abs(x[r - 1] - x[0]) != 1) {
        cout << "NO\n";
        continue;
      } else {
        t = true;
        int i = 0, j = r - 1, d1 = 1, d2 = 1;
        while ((d1 == 1 || d2 == 1) && (i + 1 <= j)) {
          d1 = abs(x[i] - x[i + 1]);
          if (d1 == 1) i += 1;
          d2 == abs(x[j] - x[j - 1]);
          if (d2 == 1) j -= 1;
        }
        if (d1 == 1 || d2 == 1 && i == j)
          cout << "YES\n";
        else
          cout << "NO\n";
      }
    }
  }
  return 0;
}
