#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
const ll mod = 1e9+7;
const ll N = 1e6 + 100;
ll spf[N];
vector<ll>V;

void sieve()
{
     bool prime[N + 1];
    memset(prime, true, sizeof(prime));
 
    for (ll p = 2; p * p <= N; p++)
    {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples
            // of p greater than or
            // equal to the square of it
            // numbers which are multiple
            // of p and are less than p^2
            // are already been marked.
            for (int i = p * p; i <= N; i += p)
                prime[i] = false;
        }
    }
 
    // Print all prime numbers
    for (ll p = 2; p <= N; p++)
        if (prime[p])
            V.push_back(p);
}

ll gcd(ll a, ll b)
{
    if(b == 0)
    {
        return a;
    }

    return gcd(b,a%b);
}

void setting()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif 

}
/*---------------------------------------------------------------------------------------------------------------------------*/


int main()
{
    setting();
    ll t;
    cin >> t;

    while(t--)
    {
        ll n,i,j;
        cin >> n;
        ll a[n];

        for(i=0; i<n; i++)
        {
            cin >> a[i];
        }

        ll ans = 2*n-1;


        for(i=0; i<=n-3; i++)
        {
            if(a[i+1] > a[i] && a[i+1] > a[i+2])
            {
                ans++;
            }
            else if(a[i+1] < a[i] && a[i+1] < a[i+2])
            {
                ans++;
            }
        }

        for(i=0; i<=n-4; i++)
        {
            if(a[i] < a[i+1] && a[i+2] < a[i] && a[i+3] > a[i+2] && a[i+3] < a[i+1])
            {
                ans++;
            }
            else if(a[i] > a[i+1] && a[i+2] > a[i] && a[i+3] < a[i+2] && a[i+3] > a[i+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
const ll mod = 1e9+7;
const ll N = 1e6 + 100;
ll spf[N];
vector<ll>V;

void sieve()
{
     bool prime[N + 1];
    memset(prime, true, sizeof(prime));
 
    for (ll p = 2; p * p <= N; p++)
    {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples
            // of p greater than or
            // equal to the square of it
            // numbers which are multiple
            // of p and are less than p^2
            // are already been marked.
            for (int i = p * p; i <= N; i += p)
                prime[i] = false;
        }
    }
 
    // Print all prime numbers
    for (ll p = 2; p <= N; p++)
        if (prime[p])
            V.push_back(p);
}

ll gcd(ll a, ll b)
{
    if(b == 0)
    {
        return a;
    }

    return gcd(b,a%b);
}

void setting()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif 

}
/*---------------------------------------------------------------------------------------------------------------------------*/


int main()
{
    setting();
    ll t;
    cin >> t;

    while(t--)
    {
        ll n,i,j;
        cin >> n;
        ll a[n];

        for(i=0; i<n; i++)
        {
            cin >> a[i];
        }

        ll ans = 2*n-1;


        for(i=0; i<=n-3; i++)
        {
            if(a[i+1] > a[i] && a[i+1] > a[i+2])
            {
                ans++;
            }
            else if(a[i+1] < a[i] && a[i+1] < a[i+2])
            {
                ans++;
            }
        }

        for(i=0; i<=n-4; i++)
        {
            if(a[i] < a[i+1] && a[i+2] < a[i] && a[i+3] > a[i+2] && a[i+3] < a[i+1])
            {
                ans++;
            }
            else if(a[i] > a[i+1] && a[i+2] > a[i] && a[i+3] < a[i+2] && a[i+3] > a[i+1])
            {
                ans++;
            }
        }

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