#include <bits/stdc++.h>
using namespace std;
int ROW[] = {+1, -1, +0, +0};
int COL[] = {+0, +0, +1, -1};
int X[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int Y[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int KX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int KY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
bool CMP(int a, int b) { return a > b; }
int a[210], n;
bool forward(int i) {
  int prev = a[i], k = 1;
  for (int j = i + 1; k < n; j++, k++) {
    if (j > n) j = 1;
    if (a[j] - prev != 1) {
      return false;
    }
    prev = a[j];
  }
  return true;
}
bool rev(int i) {
  int prev = a[i], k = 1;
  for (int j = i - 1; k < n; j--, k++) {
    if (j == 0) j = n;
    if (a[j] - prev != 1) {
      return false;
    }
    prev = a[j];
  }
  return true;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      if (forward(i)) {
        cout << "YES" << endl;
        goto ret;
      } else if (rev(i)) {
        cout << "YES" << endl;
        goto ret;
      }
    }
    cout << "NO" << endl;
  ret:;
  }
END:
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int ROW[] = {+1, -1, +0, +0};
int COL[] = {+0, +0, +1, -1};
int X[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int Y[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int KX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int KY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
bool CMP(int a, int b) { return a > b; }
int a[210], n;
bool forward(int i) {
  int prev = a[i], k = 1;
  for (int j = i + 1; k < n; j++, k++) {
    if (j > n) j = 1;
    if (a[j] - prev != 1) {
      return false;
    }
    prev = a[j];
  }
  return true;
}
bool rev(int i) {
  int prev = a[i], k = 1;
  for (int j = i - 1; k < n; j--, k++) {
    if (j == 0) j = n;
    if (a[j] - prev != 1) {
      return false;
    }
    prev = a[j];
  }
  return true;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      if (forward(i)) {
        cout << "YES" << endl;
        goto ret;
      } else if (rev(i)) {
        cout << "YES" << endl;
        goto ret;
      }
    }
    cout << "NO" << endl;
  ret:;
  }
END:
  return 0;
}

```