#include<bits/stdc++.h>
#define int ll
#define sz(x) int((x).size())
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
using pi = pair<int,int>;
const int inf = 0x3f3f3f3f3f3f3f3f;
const int minf = 0xc0c0c0c0c0c0c0c0;





signed main() {
    ios::sync_with_stdio(0); cin.tie(0);

    int TC; cin>>TC;
    for (int tc=1; tc<=TC; tc++) {
        int n; cin>>n;
        vector<int> A(n+5);

        for (int i=1; i<=n; i++) cin>>A[i];

        int res = 2*n - 1;

        for (int i=1; i<=n-2; i++) {
            vector<int> v = {A[i], A[i+1]};

            for (int j=i+2; j<=min(i+5, n); j++) {
                v.push_back(A[j]);

                bool ok = 1;

                for (int k=0; k<sz(v); k++) {
                    for (int l=k+1; l<sz(v); l++) {
                        for (int o=l+1; o<sz(v); o++) {
                            int x = (v[o] - v[l]) * (v[l] - v[k]);
                            if (x >= 0) {
                                ok = 0;
                                break;
                            }
                        }
                        if (ok == 0) break;
                    }
                    if (ok == 0) break;
                }

                if (ok == 0) break;

                ++res;
            }
        }

        cout<<res<<'\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<bits/stdc++.h>
#define int ll
#define sz(x) int((x).size())
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
using pi = pair<int,int>;
const int inf = 0x3f3f3f3f3f3f3f3f;
const int minf = 0xc0c0c0c0c0c0c0c0;





signed main() {
    ios::sync_with_stdio(0); cin.tie(0);

    int TC; cin>>TC;
    for (int tc=1; tc<=TC; tc++) {
        int n; cin>>n;
        vector<int> A(n+5);

        for (int i=1; i<=n; i++) cin>>A[i];

        int res = 2*n - 1;

        for (int i=1; i<=n-2; i++) {
            vector<int> v = {A[i], A[i+1]};

            for (int j=i+2; j<=min(i+5, n); j++) {
                v.push_back(A[j]);

                bool ok = 1;

                for (int k=0; k<sz(v); k++) {
                    for (int l=k+1; l<sz(v); l++) {
                        for (int o=l+1; o<sz(v); o++) {
                            int x = (v[o] - v[l]) * (v[l] - v[k]);
                            if (x >= 0) {
                                ok = 0;
                                break;
                            }
                        }
                        if (ok == 0) break;
                    }
                    if (ok == 0) break;
                }

                if (ok == 0) break;

                ++res;
            }
        }

        cout<<res<<'\n';
    }

    return 0;
}
```