#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include<math.h>
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tt; 
    cin >> tt; 
    while (tt--) {
        int n; 
        cin >> n; 
        vector<int> a(n), b(n); 
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        int sum = 0;
        for (int i = 2; i < n; ++i) {
            if (a[i - 1] > max(a[i], a[i - 2]) || a[i - 1] < min(a[i - 2], a[i])) {
                b[i] = 1; 
                sum += 1; 
                if (b[i - 1] == 1 && (a[i - 1] > max(a[i], a[i - 3]) || a[i - 1] < min(a[i - 3], a[i]))) {
                    int flag = 0; 
                    if (!(a[i - 1] > max(a[i], a[i - 3]) || a[i - 1] < min(a[i - 3], a[i]))) {
                        flag = 1; 
                    }
                    if (!(a[i - 2] > max(a[i], a[i - 3]) || a[i - 2] < min(a[i - 3], a[i]))) {
                        flag = 1;
                    }
                    if (flag == 0)
                        sum += 1; 
                }

            }
        }
        sum += 2 * n - 1; 
        cout << sum << "\n"; 

    }
    return 0;
    
}
    

### 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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include<math.h>
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tt; 
    cin >> tt; 
    while (tt--) {
        int n; 
        cin >> n; 
        vector<int> a(n), b(n); 
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        int sum = 0;
        for (int i = 2; i < n; ++i) {
            if (a[i - 1] > max(a[i], a[i - 2]) || a[i - 1] < min(a[i - 2], a[i])) {
                b[i] = 1; 
                sum += 1; 
                if (b[i - 1] == 1 && (a[i - 1] > max(a[i], a[i - 3]) || a[i - 1] < min(a[i - 3], a[i]))) {
                    int flag = 0; 
                    if (!(a[i - 1] > max(a[i], a[i - 3]) || a[i - 1] < min(a[i - 3], a[i]))) {
                        flag = 1; 
                    }
                    if (!(a[i - 2] > max(a[i], a[i - 3]) || a[i - 2] < min(a[i - 3], a[i]))) {
                        flag = 1;
                    }
                    if (flag == 0)
                        sum += 1; 
                }

            }
        }
        sum += 2 * n - 1; 
        cout << sum << "\n"; 

    }
    return 0;
    
}
    

```