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;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
bool cmp1(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second > b.second;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long>> v(n);
  map<pair<long long, long long>, long long> mp;
  map<pair<long long, long long>, long long> mpsort;
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    mp[{v[i].first, v[i].second}] = i + 1;
  }
  vector<long long> vis(205, 0);
  set<long long> se;
  for (long long i = 0; i < n; i++)
    for (long long j = v[i].first; j <= v[i].second; j++) {
      vis[j]++;
    }
  sort(v.begin(), v.end(), cmp);
  long long cnt = 0;
  for (auto x : v) mp[{x.first, x.second}] = cnt, cnt++;
  for (long long i = 1; i <= 200; i++)
    if (vis[i] > k) se.insert(i);
  queue<long long> q;
  for (long long x : se) {
    long long p = vis[x];
    p = p - k;
    vector<pair<long long, long long>> cur;
    for (long long i = 0; i < n; i++) {
      if (v[i].first > x)
        break;
      else if (x <= v[i].second) {
        cur.push_back({v[i].first, v[i].second});
      }
    }
    sort(cur.begin(), cur.end(), cmp1);
    for (long long i = 0; i < p; i++) {
      long long second = cur[i].first;
      long long e = cur[i].second;
      cout << second << " " << e << endl;
      for (long long j = second; j <= e; j++) {
        if (vis[j] == k + 1 && j > x) {
          se.erase(j);
        }
        vis[j]--;
      }
      long long z = mp[{cur[i].first, cur[i].second}];
      q.push(z);
    }
  }
  cout << q.size() << endl;
  while (!q.empty()) cout << q.front() << " ", q.pop();
  cout << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
