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;
void p(vector<int> v) {
  cout << "{ ";
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " , ";
  }
  cout << " }";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  vector<string> ans;
  for (int i = 0; i < q; i++) {
    int n;
    cin >> n;
    vector<int> v;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      v.push_back(x);
    }
    int a = find(v.begin(), v.end(), n) - v.begin();
    vector<int> q = v;
    sort(q.begin(), q.end());
    if (q == v) {
      ans.push_back("YES");
    } else if (v[a - 1] == 1) {
      for (int j = a; j < n; j++) {
        v.insert(v.begin() + (j - a), v[j]);
        v.erase(v.begin() + j + 1);
      }
      int c = 0;
      for (int k = 0; k < n - 1; k++) {
        if (v[k] == v[k + 1] + 1) {
          c += 1;
        }
      }
      if (c == n - 1) {
        ans.push_back("YES");
      } else {
        ans.push_back("NO");
      }
    } else if (v[a + 1] == 1) {
      for (int j = a + 1; j < n; j++) {
        v.insert(v.begin() + (j - a - 1), v[j]);
        v.erase(v.begin() + j + 1);
      }
      int b = 0;
      for (int k = 0; k < n - 1; k++) {
        if (v[k] == v[k + 1] - 1) {
          b += 1;
        }
      }
      if (b == n - 1) {
        ans.push_back("YES");
      } else {
        ans.push_back("NO");
      }
    } else {
      ans.push_back("NO");
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
}
