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 int inf = 1000000000;
const long long mo = 1e9 + 7;
long long sx, sy, ex, ey, dx[6] = {0, 1, 0, -1, 0, 0},
                          dy[6] = {1, 0, -1, 0, 0, 0}, m, n, k,
                          dz[6]{0, 0, 0, 0, -1, 1};
long long p, sg, no, v, re, ans, w;
int par[50000];
long long a[400005];
long long tr[800000][26];
long long b[400005];
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
set<char> se, se2;
long long qu(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans % m * a % m;
    }
    b >>= 1;
    a = a % m * a % m;
  }
  return ans;
}
int su(int n) {
  if (n == 1 || n == 0) return 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int fi(int a) {
  if (a == par[a])
    return a;
  else
    return par[a] = fi(par[a]);
}
vector<pair<string, string> > ve, ve1;
vector<long long> ve2;
string s1[100005];
void in(string s) {}
int cmp(int a, int b) { return a > b; }
map<long long, queue<string>[5]> mp;
int vis[200005];
int main() {
  int t, p2, p3;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, ss, sss;
  char c;
  long long l, r;
  while (cin >> t) {
    while (t--) {
      cin >> n;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
      }
      sg = 1;
      int sg2 = 1;
      for (int i = 1; i < n; i++) {
        if (a[i] - a[i - 1] != 1) {
          if (a[i] - a[i - 1] != 1 - n) {
            sg = 0;
          }
        }
      }
      for (int i = 1; i < n; i++) {
        if (a[i] - a[i - 1] != -1) {
          if (a[i] - a[i - 1] != n - 1) {
            sg2 = 0;
          }
        }
      }
      if (sg || sg2) {
        cout << "YES" << endl;
      } else
        cout << "NO" << endl;
    }
  }
}
