#include "bits/stdc++.h"
using namespace std;

typedef long long int ll;
typedef vector<ll> vi;
typedef map<ll,ll> mii;
#define pb push_back
#define mp make_pair
#define mod 998244353


bool check(ll a,ll b,ll c){
    if(b>=a and b<=c)
        return false;

    if(b>=c and b<=a)
        return false;

    return true;
}

int main()
{

#ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input1.txt", "r",stdin);
    //for writing output to output.txt
    freopen("output1.txt", "w", stdout);
#endif

ll t;
cin>>t;
    while (t--){
        
        ll n;
        cin>>n;
        ll a[n];
        for(ll i=0;i<n;i++){
            cin>>a[i];
        }    
        ll ans=0;
        ans=2*n-1;
        // for 3elements
        for(ll i=0;i<n-2;i++){
            if(check(a[i],a[i+1],a[i+2]))
                ans++;
        }
        // for 4 elements
        for(ll i=0;i<n-3;i++){
            if(check(a[i],a[i+1],a[i+2]) and check(a[i],a[i+1],a[i+3]) and check(a[i],a[i+2],a[i+3]) and check(a[i+1],a[i+2],a[i+3])){
            ans++;
            // cout<<a[i]<<" "<<a[i+1]<<" "<<a[i+2]<<" "<<a[i+3]<<endl;
        }

        }
        cout<<ans<<endl;   

    }

}






### Prompt

Your challenge is to write 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"
using namespace std;

typedef long long int ll;
typedef vector<ll> vi;
typedef map<ll,ll> mii;
#define pb push_back
#define mp make_pair
#define mod 998244353


bool check(ll a,ll b,ll c){
    if(b>=a and b<=c)
        return false;

    if(b>=c and b<=a)
        return false;

    return true;
}

int main()
{

#ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input1.txt", "r",stdin);
    //for writing output to output.txt
    freopen("output1.txt", "w", stdout);
#endif

ll t;
cin>>t;
    while (t--){
        
        ll n;
        cin>>n;
        ll a[n];
        for(ll i=0;i<n;i++){
            cin>>a[i];
        }    
        ll ans=0;
        ans=2*n-1;
        // for 3elements
        for(ll i=0;i<n-2;i++){
            if(check(a[i],a[i+1],a[i+2]))
                ans++;
        }
        // for 4 elements
        for(ll i=0;i<n-3;i++){
            if(check(a[i],a[i+1],a[i+2]) and check(a[i],a[i+1],a[i+3]) and check(a[i],a[i+2],a[i+3]) and check(a[i+1],a[i+2],a[i+3])){
            ans++;
            // cout<<a[i]<<" "<<a[i+1]<<" "<<a[i+2]<<" "<<a[i+3]<<endl;
        }

        }
        cout<<ans<<endl;   

    }

}






```