#include<bits/stdc++.h>
using namespace std;
 
#define ll long long 
#define F first 
#define S second
const ll MOD=998244353;

const ll MX=1e5+7;
const int sz=200;
vector<int> parent;

void initialize(){   
}


bool check(vector<int> a){
    int n=a.size();
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<n;k++){
                // cout<<i<<' '<<j<<' '<<k<<"\n";
                if(i==k or j==k or i==j)
                    continue;
                // cout<<i<<' '<<j<<' '<<k<<"\n";
                if(abs(j-i)+abs(k-j)+abs(a[j]-a[i])+abs(a[k]-a[j])==abs(k-i)+abs(a[k]-a[i])){
                    // cout<<a[i]<<' '<<a[j]<<' '<<a[k]<<'\n';
                    return false;
                }

            }
        }
    }
    return true;
}

void test_case(int t){
    int n;cin>>n;
    vector<int> arr(n);
    for(int &x:arr)cin>>x;

    int ans=n+n-1;
    for(int i=0;i<n-2;i++){
        ans+=check({arr[i],arr[i+1],arr[i+2]});
        if(i<n-3){
            ans+=check({arr[i],arr[i+1],arr[i+2],arr[i+3]});
        }
    }
    cout<<ans<<'\n';

}


    
 
int main(){
    initialize();
    int t;//t=1;
    cin>>t;
    while(t--){
        test_case(t+1);
    }
}
 
    
### Prompt

Your challenge is to write a cpp solution to 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>
using namespace std;
 
#define ll long long 
#define F first 
#define S second
const ll MOD=998244353;

const ll MX=1e5+7;
const int sz=200;
vector<int> parent;

void initialize(){   
}


bool check(vector<int> a){
    int n=a.size();
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<n;k++){
                // cout<<i<<' '<<j<<' '<<k<<"\n";
                if(i==k or j==k or i==j)
                    continue;
                // cout<<i<<' '<<j<<' '<<k<<"\n";
                if(abs(j-i)+abs(k-j)+abs(a[j]-a[i])+abs(a[k]-a[j])==abs(k-i)+abs(a[k]-a[i])){
                    // cout<<a[i]<<' '<<a[j]<<' '<<a[k]<<'\n';
                    return false;
                }

            }
        }
    }
    return true;
}

void test_case(int t){
    int n;cin>>n;
    vector<int> arr(n);
    for(int &x:arr)cin>>x;

    int ans=n+n-1;
    for(int i=0;i<n-2;i++){
        ans+=check({arr[i],arr[i+1],arr[i+2]});
        if(i<n-3){
            ans+=check({arr[i],arr[i+1],arr[i+2],arr[i+3]});
        }
    }
    cout<<ans<<'\n';

}


    
 
int main(){
    initialize();
    int t;//t=1;
    cin>>t;
    while(t--){
        test_case(t+1);
    }
}
 
    
```