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;
const long long mod = 1e9 + 7;
long long xor1_n(long long n) {
  switch (n % 4) {
    case 0:
      return n;
    case 1:
      return 1;
    case 2:
      return n + 1;
  }
  return 0;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
clock_t time_p = clock();
void rtime() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long power(long long a, long long b, long long m = mod) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % m;
    b >>= 1;
    a = (1ll * a * a) % m;
  }
  return ans;
}
long long rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    long long x;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] == 1) x = i;
    }
    cerr << x << " ";
    long long k = 1;
    for (long long i = x; i < n; i++)
      if (a[i] == k) k++;
    for (long long i = 0; i < x; i++)
      if (a[i] == k) k++;
    if (k == n + 1) {
      cout << "YES\n";
      continue;
    }
    k = 1;
    for (long long i = x; i >= 0; i--)
      if (a[i] == k) k++;
    for (long long i = n - 1; i > x; i--)
      if (a[i] == k) k++;
    if (k == n + 1) {
      cout << "YES\n";
      continue;
    }
    cout << "NO\n";
  }
  rtime();
  return 0;
}
