#include <bits/stdc++.h>
using namespace std;
#define rep(i,j,n) for(int i=j;i<=n;i++)
#define repp(i,j,n) for(int i=n;i>=j;i--)
#define pb push_back
#define in insert
#define ll long long int
#define pii pair<int,int>
#define pip pair<int,pii>
#define pll pair<ll,ll>
#define pib pair<int,bool>
#define p_q priority_queue
#define mp map<int,int>
#define mpl map<ll,ll>
#define inf 1000000005
# define fi first 
# define se second 
#define mod 1000000007
#define vi vector<int>
#define vpi vector<pi>
#define vll vector<ll>
#define qu queue
#define N 1000005
#define BLK 700
#define pi 3.1415926535897932384626433832795
#define lowercase(a) transform(a.begin(), a.end(), a.begin(), ::tolower);
#define uppercase(a) transform(a.begin(), a.end(), a.begin(), ::toupper);
int main()
{
ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;cin>>t;
    while(t--){
        ll n;
        cin>>n;
        ll a[n];
        rep(i,0,n-1) cin>>a[i];
        ll ans = n + n-1;
        int w,x,y,z;
        rep(i,0,n-3){
        x=a[i];y=a[i+1];z=a[i+2];
        if(x>=y && y>=z || x<=y && y<=z){}
        else{ans++;}
        }
        //cout<<ans<<endl;
        rep(i,0,n-4){
        x=a[i];y=a[i+1];z=a[i+2];w=a[i+3];
        if(x>=y && y>=z || x<=y && y<=z || x>=y && y>=w || x<=y && y<=w || y>=z && z>=w || y<=z && z<=w || x>=z && z>=w || x<=z && z<=w){}
        else{ans++;}
        }
        cout<<ans<<endl;
    }
    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>
using namespace std;
#define rep(i,j,n) for(int i=j;i<=n;i++)
#define repp(i,j,n) for(int i=n;i>=j;i--)
#define pb push_back
#define in insert
#define ll long long int
#define pii pair<int,int>
#define pip pair<int,pii>
#define pll pair<ll,ll>
#define pib pair<int,bool>
#define p_q priority_queue
#define mp map<int,int>
#define mpl map<ll,ll>
#define inf 1000000005
# define fi first 
# define se second 
#define mod 1000000007
#define vi vector<int>
#define vpi vector<pi>
#define vll vector<ll>
#define qu queue
#define N 1000005
#define BLK 700
#define pi 3.1415926535897932384626433832795
#define lowercase(a) transform(a.begin(), a.end(), a.begin(), ::tolower);
#define uppercase(a) transform(a.begin(), a.end(), a.begin(), ::toupper);
int main()
{
ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;cin>>t;
    while(t--){
        ll n;
        cin>>n;
        ll a[n];
        rep(i,0,n-1) cin>>a[i];
        ll ans = n + n-1;
        int w,x,y,z;
        rep(i,0,n-3){
        x=a[i];y=a[i+1];z=a[i+2];
        if(x>=y && y>=z || x<=y && y<=z){}
        else{ans++;}
        }
        //cout<<ans<<endl;
        rep(i,0,n-4){
        x=a[i];y=a[i+1];z=a[i+2];w=a[i+3];
        if(x>=y && y>=z || x<=y && y<=z || x>=y && y>=w || x<=y && y<=w || y>=z && z>=w || y<=z && z<=w || x>=z && z>=w || x<=z && z<=w){}
        else{ans++;}
        }
        cout<<ans<<endl;
    }
    return 0;
}

```