#include <bits/stdc++.h>

using namespace std;

#define ar array
#define ll long long
typedef int uci;
#define int long long
#define F first
#define S second

const int MAX_N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;

//! function insert

//THINK FIRST, CODE SECOND
//DON'T GET STUCK ON ONE STRATEGY
//CALM DOWNNN FOR ONCE IN YOUR LIFE
//REDUCE
//COUGH E??!?!?!! O.O
//uwu i see you ryan

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i{}; i < n; ++i)
        cin >> a[i];

    int out = n + n -1;
    vector<bool> three(n);
    for(int i{1}; i < n-1; ++i){
        if(a[i] > a[i-1] && a[i+1] < a[i]){
            three[i] = a[i+1] < a[i-1];
            out++;
        }else if(a[i] < a[i-1] && a[i+1] > a[i]){
            out++;
            three[i] = a[i+1] > a[i-1];
        }
    }
    for(int i{1}; i < n-2; ++i){
        if(!three[i])
            continue;
        if(a[i+1] < a[i] && a[i+2] > a[i+1] && a[i+2] < a[i] && a[i-1] != a[i+1] && a[i+2] != a[i])
            out++;
        else if(a[i+1] > a[i] && a[i+2] < a[i+1] && a[i+2] > a[i] && a[i-1] != a[i+1] && a[i+2] != a[i])
            out++;
    }
    cout << out << '\n';
}

uci main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int tc; cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}
### 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;

#define ar array
#define ll long long
typedef int uci;
#define int long long
#define F first
#define S second

const int MAX_N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;

//! function insert

//THINK FIRST, CODE SECOND
//DON'T GET STUCK ON ONE STRATEGY
//CALM DOWNNN FOR ONCE IN YOUR LIFE
//REDUCE
//COUGH E??!?!?!! O.O
//uwu i see you ryan

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i{}; i < n; ++i)
        cin >> a[i];

    int out = n + n -1;
    vector<bool> three(n);
    for(int i{1}; i < n-1; ++i){
        if(a[i] > a[i-1] && a[i+1] < a[i]){
            three[i] = a[i+1] < a[i-1];
            out++;
        }else if(a[i] < a[i-1] && a[i+1] > a[i]){
            out++;
            three[i] = a[i+1] > a[i-1];
        }
    }
    for(int i{1}; i < n-2; ++i){
        if(!three[i])
            continue;
        if(a[i+1] < a[i] && a[i+2] > a[i+1] && a[i+2] < a[i] && a[i-1] != a[i+1] && a[i+2] != a[i])
            out++;
        else if(a[i+1] > a[i] && a[i+2] < a[i+1] && a[i+2] > a[i] && a[i-1] != a[i+1] && a[i+2] != a[i])
            out++;
    }
    cout << out << '\n';
}

uci main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int tc; cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}
```