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

typedef vector<int> vi;
typedef pair<int, int> pii;
#define endl "\n"
#define sd(val) scanf("%d", &val)
#define ss(val) scanf("%s", &val)
#define sl(val) scanf("%lld", &val)
#define debug(val) printf("check%d\n", val)
#define all(v) v.begin(), v.end()
#define PB push_back
#define MP make_pair
#define FF first
#define SS second
#define ll long long
#define MOD 1000000007
#define clr(val) memset(val, 0, sizeof(val))
#define what_is(x) cerr << #x << " is " << x << endl;
#define OJ                            \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);
#define FIO                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    auto isGood = [&](int l, int r)
    {
        for (int i = l; i <= r; i++)
        {
            for (int j = i + 1; j<= r; j++)
            {
                for (int k = j + 1; k <= r; k++)
                {
                    if (a[i] <= a[j] && a[j] <= a[k])
                        return false;
                    if (a[i] >= a[j] && a[j] >= a[k])
                        return false;
                }
            }
        }
        return true;
    };
    int ans = 0;
    for(int i =0, j = 0; i<n;i++){
        while(!isGood(j,i)){
            j++;
        }
        ans+= i-j+1;
    }
    cout<<ans<<endl;

}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 vector<int> vi;
typedef pair<int, int> pii;
#define endl "\n"
#define sd(val) scanf("%d", &val)
#define ss(val) scanf("%s", &val)
#define sl(val) scanf("%lld", &val)
#define debug(val) printf("check%d\n", val)
#define all(v) v.begin(), v.end()
#define PB push_back
#define MP make_pair
#define FF first
#define SS second
#define ll long long
#define MOD 1000000007
#define clr(val) memset(val, 0, sizeof(val))
#define what_is(x) cerr << #x << " is " << x << endl;
#define OJ                            \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);
#define FIO                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    auto isGood = [&](int l, int r)
    {
        for (int i = l; i <= r; i++)
        {
            for (int j = i + 1; j<= r; j++)
            {
                for (int k = j + 1; k <= r; k++)
                {
                    if (a[i] <= a[j] && a[j] <= a[k])
                        return false;
                    if (a[i] >= a[j] && a[j] >= a[k])
                        return false;
                }
            }
        }
        return true;
    };
    int ans = 0;
    for(int i =0, j = 0; i<n;i++){
        while(!isGood(j,i)){
            j++;
        }
        ans+= i-j+1;
    }
    cout<<ans<<endl;

}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}
```