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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
inline ll getint() {
  ll x = 0, s = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') c = getchar(), s = -1;
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * s;
}
const ll MOD7 = 1e9 + 7;
const ll MAX = 1e6;
const ll INF = 1023456789;
inline ll fpow(ll a, ll b, ll m = MOD7) {
  if (b == 0) return 1 % m;
  if (b == 1) return a % m;
  ll r = fpow((a * a) % m, b / 2, m);
  return (r * (b & 1 ? a : 1)) % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(NULL) * time(NULL) ^ hash<ll>{}(INF * time(NULL) % MOD7));
  int N, dx, start;
  vector<int> P;
  unordered_set<int> s;
  int Q;
  cin >> Q;
  while (Q--) {
    cin >> N;
    P.resize(N);
    for (auto i = (0); i < (N); ++i) cin >> P[i];
    s.clear();
    for (auto i = (0); i < (N - 1); ++i) s.insert(P[i + 1] - P[i]);
    s.insert(P[0] - P[N - 1]);
    cout << (s.size() <= 2 && (s.find(1) != s.end() || s.find(-1) != s.end()) &&
                     s.find(N - 1) != s.end()
                 ? "YES"
                 : "NO")
         << '\n';
  }
  return 0;
}
