#include<bits/stdc++.h>
#include <iterator>
#include <iostream>
#include <numeric>
#include <math.h>
#define ll long long
#define ull long
#define mpa make_pair
#define pb push_back
#define ff first
#define pii pair<ll,ll>
#define dd double
#define trace(x) cerr << #x << " : " << x << endl
#define ss second
#define boost ios_base::sync_with_stdio(0)
#define forp(i,a,b) for(ll i=a;i<=b;i++)
#define rep(i,n)    for(ll i=0;i<n;++i)
#define ren(i,n)    for(ll i=n-1;i>=0;i--)
#define forn(i,a,b) for(ll i=a;i>=b;i--)
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(typeof((c).begin()) i = (c).begin(); i != (c).end();
#define sc(x) scanf("%lld",&x)
#define clr(x,val) memset(x,val,sizeof(x))
#define pr(x) printf("%lld\n",x) 
#define gc getchar
#define pdd pair<dd,dd>
#define prec(x) cout<<fixed<<setprecision(x)
#define fre freopen("input.txt","r",stdin),freopen("output.txt","w",stdout)
#define arr array 
using namespace std;

ll a[200005];
ll dist(ll x,ll y){
    return abs(x-y)+abs(a[x]-a[y]);
}
void solve(ll tc) {
    ll n;
    cin>>n;
    for(ll i=1;i<=n;i++){
       cin>>a[i];
    }
    ll ans=0;

    for(ll i=1;i<=n;i++){
        ll end=n;

        for(ll j=i+1;j<=n;j++){
            for(ll x=i;x<j;x++){
                for(ll y=x+1;y<j;y++){
                    if(dist(x,y)+dist(y,j)==dist(x,j)){
                        end=j-1;
                        break;
                    }
                }
                if(end !=n){
                    break;
                }
            }

            if(end !=n){
                    break;
            }
            
        }

        //cout<<i<<" "<<end<<endl;         
        ans+=end-i+1;
    }
       



    cout<<ans<<endl;





   
}
 
int main(){
    
   
	boost;
    
    
	ll t=1;
    ll tc=1;
	cin>>t;

	while(t--){
		solve(tc);
        tc++;
	}

    return 0;
    
     
}
  
### Prompt

Please create a solution in CPP 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>
#include <iterator>
#include <iostream>
#include <numeric>
#include <math.h>
#define ll long long
#define ull long
#define mpa make_pair
#define pb push_back
#define ff first
#define pii pair<ll,ll>
#define dd double
#define trace(x) cerr << #x << " : " << x << endl
#define ss second
#define boost ios_base::sync_with_stdio(0)
#define forp(i,a,b) for(ll i=a;i<=b;i++)
#define rep(i,n)    for(ll i=0;i<n;++i)
#define ren(i,n)    for(ll i=n-1;i>=0;i--)
#define forn(i,a,b) for(ll i=a;i>=b;i--)
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(typeof((c).begin()) i = (c).begin(); i != (c).end();
#define sc(x) scanf("%lld",&x)
#define clr(x,val) memset(x,val,sizeof(x))
#define pr(x) printf("%lld\n",x) 
#define gc getchar
#define pdd pair<dd,dd>
#define prec(x) cout<<fixed<<setprecision(x)
#define fre freopen("input.txt","r",stdin),freopen("output.txt","w",stdout)
#define arr array 
using namespace std;

ll a[200005];
ll dist(ll x,ll y){
    return abs(x-y)+abs(a[x]-a[y]);
}
void solve(ll tc) {
    ll n;
    cin>>n;
    for(ll i=1;i<=n;i++){
       cin>>a[i];
    }
    ll ans=0;

    for(ll i=1;i<=n;i++){
        ll end=n;

        for(ll j=i+1;j<=n;j++){
            for(ll x=i;x<j;x++){
                for(ll y=x+1;y<j;y++){
                    if(dist(x,y)+dist(y,j)==dist(x,j)){
                        end=j-1;
                        break;
                    }
                }
                if(end !=n){
                    break;
                }
            }

            if(end !=n){
                    break;
            }
            
        }

        //cout<<i<<" "<<end<<endl;         
        ans+=end-i+1;
    }
       



    cout<<ans<<endl;





   
}
 
int main(){
    
   
	boost;
    
    
	ll t=1;
    ll tc=1;
	cin>>t;

	while(t--){
		solve(tc);
        tc++;
	}

    return 0;
    
     
}
  
```