#include <bits/stdc++.h>
#define maxn (ll)(2e5+5)
typedef long long ll;
using namespace std;
ll a[maxn]={0};

void solve()
{
    ll n,ans=0;
    cin>>n;
    for(int i=1;i<=n;++i)
    {
        scanf("%lld",&a[i]);
    }
    ans+=2*n-1;
    for(int i=1;i+2<=n;i++)
    {
        if(i==n-2)
        {
            if((a[i]>a[i+2] && a[i+2]>a[i+1])||
               (a[i+2]>a[i] && a[i]>a[i+1])  ||
               ((a[i+1]>a[i+2])&&(a[i+2]>a[i]))||
               ((a[i+1]>a[i])&&(a[i]>a[i+2]))||
               ((a[i]==a[i+2])&&(a[i+1]!=a[i])))
                {ans++;}
        }
        else
        {
            if((a[i]>a[i+2] && a[i+2]>a[i+1]) ||((a[i]==a[i+2])&&(a[i+1]!=a[i]))||(a[i+1]>a[i+2] && a[i+2]>a[i]))
            {
                ans++;
            }
            if(a[i+2]>a[i] && a[i]>a[i+1])
            {
                ans++;
                if(a[i+3]<a[i+2] && a[i+3]>a[i+1])
                    {ans++;}
            }
            if(a[i+1]>a[i] && a[i]>a[i+2])
            {
                ans++;
                if(a[i+3]>a[i+2] && a[i+3]<a[i+1])
                    {ans++;}
            }
        }
    }
    cout<<ans<<endl;

}


int main()
{
  int t;
  cin>>t;
  while(t--) solve();
  return 0;
}

### Prompt

Please formulate a CPP solution to 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>
#define maxn (ll)(2e5+5)
typedef long long ll;
using namespace std;
ll a[maxn]={0};

void solve()
{
    ll n,ans=0;
    cin>>n;
    for(int i=1;i<=n;++i)
    {
        scanf("%lld",&a[i]);
    }
    ans+=2*n-1;
    for(int i=1;i+2<=n;i++)
    {
        if(i==n-2)
        {
            if((a[i]>a[i+2] && a[i+2]>a[i+1])||
               (a[i+2]>a[i] && a[i]>a[i+1])  ||
               ((a[i+1]>a[i+2])&&(a[i+2]>a[i]))||
               ((a[i+1]>a[i])&&(a[i]>a[i+2]))||
               ((a[i]==a[i+2])&&(a[i+1]!=a[i])))
                {ans++;}
        }
        else
        {
            if((a[i]>a[i+2] && a[i+2]>a[i+1]) ||((a[i]==a[i+2])&&(a[i+1]!=a[i]))||(a[i+1]>a[i+2] && a[i+2]>a[i]))
            {
                ans++;
            }
            if(a[i+2]>a[i] && a[i]>a[i+1])
            {
                ans++;
                if(a[i+3]<a[i+2] && a[i+3]>a[i+1])
                    {ans++;}
            }
            if(a[i+1]>a[i] && a[i]>a[i+2])
            {
                ans++;
                if(a[i+3]>a[i+2] && a[i+3]<a[i+1])
                    {ans++;}
            }
        }
    }
    cout<<ans<<endl;

}


int main()
{
  int t;
  cin>>t;
  while(t--) solve();
  return 0;
}

```