#include <bits/stdc++.h>

void solve() {
    int n;
    std::cin >> n;
    std::vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        std::cin >> a[i];
    }
    int ans = 2 * n - 1;
    for (int i = 0; i + 2 < n; ++i) {
        for (int j = i + 2; j < n; ++j) {
            bool flag = true;
            for (int x = i; x < j; ++x) {
                for (int y = x + 1; y < j; ++y) {
                    if ((a[x] <= a[y] && a[y] <= a[j]) ||
                        (a[x] >= a[y] && a[y] >= a[j])) {
                        flag = false;
                    }
                }
            }
            if (flag) {
                ++ans;
            } else {
                break;
            }
        }
    }
    std::cout << ans << "\n";
}

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);

    int T;
    std::cin >> T;
    while (T--) {
        solve();
    }
}
### Prompt

Create a solution in Cpp for 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>

void solve() {
    int n;
    std::cin >> n;
    std::vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        std::cin >> a[i];
    }
    int ans = 2 * n - 1;
    for (int i = 0; i + 2 < n; ++i) {
        for (int j = i + 2; j < n; ++j) {
            bool flag = true;
            for (int x = i; x < j; ++x) {
                for (int y = x + 1; y < j; ++y) {
                    if ((a[x] <= a[y] && a[y] <= a[j]) ||
                        (a[x] >= a[y] && a[y] >= a[j])) {
                        flag = false;
                    }
                }
            }
            if (flag) {
                ++ans;
            } else {
                break;
            }
        }
    }
    std::cout << ans << "\n";
}

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);

    int T;
    std::cin >> T;
    while (T--) {
        solve();
    }
}
```