#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define mod 1000000007
#define mo 1000000007
#define mod2 1000000000
#define mod1 998244353
#define lol ios_base::sync_with_stdio(false);cin.tie(NULL);
#define all(x) (x).begin(), (x).end()
#define F(i, n) for (ll (i) = 0; (i) < (n); (i)++)
#define pb push_back
#define pf push_front
#define mp make_pair
#define f first
#define ss second
#define MAXN   1000001

bool check(ll arr[],int n,int i,int j)
{
    for(ll ii=i;ii<=j;ii++)
    {
        for(ll jj=ii+1;jj<=j;jj++)
        {
            for(ll kk=jj+1;kk<=j;kk++)
            {
                if(arr[ii]<=arr[jj] && arr[jj]<=arr[kk]) 
                {
                    return false;
                }
                
                if(arr[ii]>=arr[jj] && arr[jj]>=arr[kk]) 
                {
                    return false;
                }
            }
        }
    }
    return true;
}
int main()
{
    int t;cin >> t;
    while(t--)
    {
        ll n;cin >> n;
        ll arr[n];
        for(ll i=0;i<n;i++) cin >> arr[i];
        if(n<=2)
        {
            cout << n + n - 1 << endl;
            continue;
        }
        
        
        ll ans = n + (n-1);
        for(ll k=3;k<=min(n,(ll)20);k++)
        {
            for(ll i=0;i+k-1<n;i++) 
            {
                if(check(arr,n,i,i+k-1)) ans++; 
            }
        }

        cout << ans << endl;
        
    }
    
}
### 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>
using namespace std;
#define ll long long
#define ld long double
#define mod 1000000007
#define mo 1000000007
#define mod2 1000000000
#define mod1 998244353
#define lol ios_base::sync_with_stdio(false);cin.tie(NULL);
#define all(x) (x).begin(), (x).end()
#define F(i, n) for (ll (i) = 0; (i) < (n); (i)++)
#define pb push_back
#define pf push_front
#define mp make_pair
#define f first
#define ss second
#define MAXN   1000001

bool check(ll arr[],int n,int i,int j)
{
    for(ll ii=i;ii<=j;ii++)
    {
        for(ll jj=ii+1;jj<=j;jj++)
        {
            for(ll kk=jj+1;kk<=j;kk++)
            {
                if(arr[ii]<=arr[jj] && arr[jj]<=arr[kk]) 
                {
                    return false;
                }
                
                if(arr[ii]>=arr[jj] && arr[jj]>=arr[kk]) 
                {
                    return false;
                }
            }
        }
    }
    return true;
}
int main()
{
    int t;cin >> t;
    while(t--)
    {
        ll n;cin >> n;
        ll arr[n];
        for(ll i=0;i<n;i++) cin >> arr[i];
        if(n<=2)
        {
            cout << n + n - 1 << endl;
            continue;
        }
        
        
        ll ans = n + (n-1);
        for(ll k=3;k<=min(n,(ll)20);k++)
        {
            for(ll i=0;i+k-1<n;i++) 
            {
                if(check(arr,n,i,i+k-1)) ans++; 
            }
        }

        cout << ans << endl;
        
    }
    
}
```