#include<bits/stdc++.h>
#define endl '\n'
#define mt make_tuple
#define pb push_back
#define ff first
#define ss second
#define ALL(v) v.begin(), v.end()


using namespace std;
using ll = long long;
using ld = long double;

const int maxn = 100001;
const ll INF = LLONG_MAX;
const ll mod = 1e9 + 7;

ll POW(ll a, ll b, ll MMM = mod) {ll ret=1; for(;b;b>>=1,a=(a*a)%MMM)if(b&1)ret=(ret*a)% MMM; return ret; }

int n;
vector<int> a;

inline bool check(int l, int r){
  for(int i = l; i <= r; i++){
    for(int j = i+1; j <= r; j++){
      for(int k = j+1; k <= r; k++){
        if((a[i] <= a[j] && a[j] <= a[k]) || (a[i] >= a[j] && a[j] >= a[k])){
          return false;
        }
      }
    }
  }

  return true;
}  

int go_vishal(){
  cin >> n;
  a = vector<int> (n);
  for(auto &x : a){
    cin >> x;
  }

  
  int j = 0;
  int ans= 0;
  for(int i = 0; i < n; i++){
    while(!check(j, i)){
      j++;
    }
    ans += i - j + 1;
  }

  cout << ans << endl;
  return 0;
}


int main(){
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int tt;
  tt = 1;
  cin >> tt;
  for(int i = 1; i <= tt; i++){
    // cout << "Case #" << i << ": ";
    go_vishal();
  }
  return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

Suppose you have two points p = (x_p, y_p) and q = (x_q, y_q). Let's denote the Manhattan distance between them as d(p, q) = |x_p - x_q| + |y_p - y_q|.

Let's say that three points p, q, r form a bad triple if d(p, r) = d(p, q) + d(q, r).

Let's say that an array b_1, b_2, ..., b_m is good if it is impossible to choose three distinct indices i, j, k such that the points (b_i, i), (b_j, j) and (b_k, k) form a bad triple.

You are given an array a_1, a_2, ..., a_n. Calculate the number of good subarrays of a. A subarray of the array a is the array a_l, a_{l + 1}, ..., a_r for some 1 ≤ l ≤ r ≤ n.

Note that, according to the definition, subarrays of length 1 and 2 are good.

Input

The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of array a.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the number of good subarrays of array a.

Example

Input


3
4
2 4 1 3
5
6 9 1 9 6
2
13 37


Output


10
12
3

Note

In the first test case, it can be proven that any subarray of a is good. For example, subarray [a_2, a_3, a_4] is good since it contains only three elements and: 

  * d((a_2, 2), (a_4, 4)) = |4 - 3| + |2 - 4| = 3 < d((a_2, 2), (a_3, 3)) + d((a_3, 3), (a_4, 4)) = 3 + 1 + 2 + 1 = 7; 
  * d((a_2, 2), (a_3, 3)) < d((a_2, 2), (a_4, 4)) + d((a_4, 4), (a_3, 3)); 
  * d((a_3, 3), (a_4, 4)) < d((a_3, 3), (a_2, 2)) + d((a_2, 2), (a_4, 4)); 



In the second test case, for example, subarray [a_1, a_2, a_3, a_4] is not good, since it contains a bad triple (a_1, 1), (a_2, 2), (a_4, 4): 

  * d((a_1, 1), (a_4, 4)) = |6 - 9| + |1 - 4| = 6; 
  * d((a_1, 1), (a_2, 2)) = |6 - 9| + |1 - 2| = 4; 
  * d((a_2, 2), (a_4, 4)) = |9 - 9| + |2 - 4| = 2; 



So, d((a_1, 1), (a_4, 4)) = d((a_1, 1), (a_2, 2)) + d((a_2, 2), (a_4, 4)).

### Response

```cpp
#include<bits/stdc++.h>
#define endl '\n'
#define mt make_tuple
#define pb push_back
#define ff first
#define ss second
#define ALL(v) v.begin(), v.end()


using namespace std;
using ll = long long;
using ld = long double;

const int maxn = 100001;
const ll INF = LLONG_MAX;
const ll mod = 1e9 + 7;

ll POW(ll a, ll b, ll MMM = mod) {ll ret=1; for(;b;b>>=1,a=(a*a)%MMM)if(b&1)ret=(ret*a)% MMM; return ret; }

int n;
vector<int> a;

inline bool check(int l, int r){
  for(int i = l; i <= r; i++){
    for(int j = i+1; j <= r; j++){
      for(int k = j+1; k <= r; k++){
        if((a[i] <= a[j] && a[j] <= a[k]) || (a[i] >= a[j] && a[j] >= a[k])){
          return false;
        }
      }
    }
  }

  return true;
}  

int go_vishal(){
  cin >> n;
  a = vector<int> (n);
  for(auto &x : a){
    cin >> x;
  }

  
  int j = 0;
  int ans= 0;
  for(int i = 0; i < n; i++){
    while(!check(j, i)){
      j++;
    }
    ans += i - j + 1;
  }

  cout << ans << endl;
  return 0;
}


int main(){
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int tt;
  tt = 1;
  cin >> tt;
  for(int i = 1; i <= tt; i++){
    // cout << "Case #" << i << ": ";
    go_vishal();
  }
  return 0;
}


```