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 N, M, K, A, B, X, Y, Z, C, a, b, c, d, i, j, k;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pow(long long first, long long second) {
  long long res = 1;
  while (second > 0) {
    if (second & 1) res = res * first;
    second = second >> 1;
    first = first * first;
  }
  return res;
}
long long power(long long first, long long second, long long p) {
  second = second % (p - 1);
  long long res = 1;
  first = first % p;
  while (second > 0) {
    if (second & 1) res = (res * first) % p;
    second = second >> 1;
    first = (first * first) % p;
  }
  return res;
}
bool comp(string A, string B) {
  string X = A + B;
  string Y = B + A;
  if (X.compare(Y) >= 0) {
    return true;
  }
  return false;
}
void solve() {
  cin >> N;
  long long A[N];
  for (int i = 0; i < N; i++) cin >> A[i];
  bool p = 1, W = N + 1;
  for (int i = 1; i < N; i++) {
    long long r = A[i - 1] + 1;
    if (r > N) r -= N;
    if (A[i] != r) {
      p = 0;
      break;
    }
  }
  if (p == 1) {
    cout << "YES" << endl;
    return;
  }
  p = 1;
  for (int i = 1; i < N; i++) {
    long long r = A[i - 1] - 1;
    if (r == 0) r = N;
    if (A[i] != r) {
      p = 0;
      break;
    }
  }
  if (p == 1) {
    cout << "YES" << endl;
    return;
  }
  cout << "NO" << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
