#include <bits/stdc++.h>
using namespace std;
 
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp> 
//using namespace __gnu_pbds; 
//https://codeforces.com/blog/entry/11080
//#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
//std::nth_element(v, v + 4, v + 8);
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//#define endl "\n"
#define ll long long
#define f first
#define s second
#define pii pair<int,int>
#define lui long unsigned int
//#define int long long
 
const int N = 1000005;
const ll MOD = 1e9+7;
const ll M = 1e18;
 
ll qpow(ll a, ll b, ll m)
{
    ll  ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%m;
        b/=2;
        a=(a*a)%m;
    }
    return ans;
}

int main()
{
 
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    IOS;
 
    int t=1; cin>>t;
    while(t--){
      int n; cin >> n;
      int a[n];
      for(int i=0; i<n; i++)
        cin >> a[i];
      int ans = 2*n-1;
      bool ok;

      for(int i=0; i<n; i++){
        if(i+2 < n){
           ok = true;
           if(abs(a[i+2]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+2]-a[i+1]))
            ok = false;
           if(ok) ans++;
        }
        if(i+3 < n){
           ok = true;
           if(abs(a[i+2]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+2]-a[i+1]))
            ok = false;
           if(abs(a[i+3]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+3]-a[i+1]))
            ok = false;
           if(abs(a[i+3]-a[i]) == abs(a[i+2]-a[i]) + abs(a[i+3]-a[i+2]))
            ok = false;
           if(abs(a[i+3]-a[i+1]) == abs(a[i+2]-a[i+1]) + abs(a[i+3]-a[i+2]))
            ok = false;
           if(ok) ans++;
        }
      }
      cout << ans;
      
      cout << endl;
        
    }
    
    return 0;
}
### Prompt

Create a solution in cpp for the following problem:

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>
using namespace std;
 
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp> 
//using namespace __gnu_pbds; 
//https://codeforces.com/blog/entry/11080
//#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
//std::nth_element(v, v + 4, v + 8);
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//#define endl "\n"
#define ll long long
#define f first
#define s second
#define pii pair<int,int>
#define lui long unsigned int
//#define int long long
 
const int N = 1000005;
const ll MOD = 1e9+7;
const ll M = 1e18;
 
ll qpow(ll a, ll b, ll m)
{
    ll  ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%m;
        b/=2;
        a=(a*a)%m;
    }
    return ans;
}

int main()
{
 
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    IOS;
 
    int t=1; cin>>t;
    while(t--){
      int n; cin >> n;
      int a[n];
      for(int i=0; i<n; i++)
        cin >> a[i];
      int ans = 2*n-1;
      bool ok;

      for(int i=0; i<n; i++){
        if(i+2 < n){
           ok = true;
           if(abs(a[i+2]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+2]-a[i+1]))
            ok = false;
           if(ok) ans++;
        }
        if(i+3 < n){
           ok = true;
           if(abs(a[i+2]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+2]-a[i+1]))
            ok = false;
           if(abs(a[i+3]-a[i]) == abs(a[i+1]-a[i]) + abs(a[i+3]-a[i+1]))
            ok = false;
           if(abs(a[i+3]-a[i]) == abs(a[i+2]-a[i]) + abs(a[i+3]-a[i+2]))
            ok = false;
           if(abs(a[i+3]-a[i+1]) == abs(a[i+2]-a[i+1]) + abs(a[i+3]-a[i+2]))
            ok = false;
           if(ok) ans++;
        }
      }
      cout << ans;
      
      cout << endl;
        
    }
    
    return 0;
}
```