#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h> 
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
ll MOD = 998244353;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 

void solve(){
    int n;
    cin>>n;
    vector<int> arr(n);
    for(auto &e : arr) {
        cin>>e;
    }
    //ayo wuss poppin
    
    ll ans=0;
    for(int len=3; len<=4; len++) {
        if(len>n) break;
        for(int i=0; i<n-len+1; i++) {
            vector<int> subarr;
            bool ok=1;
            for(int j=i; j<len+i; j++) {
                subarr.pb(arr[j]);
            }
            for(int i=0; i<len; i++) {
                for(int j=i+1; j<len; j++) {
                    for(int k=j+1; k<len; k++) {
                        if(subarr[i]<=subarr[j]&&subarr[j]<=subarr[k]) {
                            ok=0;
                        }
                        if(subarr[i]>=subarr[j]&&subarr[j]>=subarr[k]) {
                            ok=0;
                        }
                    }
                }
            }
            if(ok) {
                ans++;
            }
        }

    }
    ans+=2*n-1;
    cout<<ans<<endl;
}
int main()
{
    fast_cin();
    int t;
    cin>>t;
    while(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
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h> 
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
ll MOD = 998244353;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 

void solve(){
    int n;
    cin>>n;
    vector<int> arr(n);
    for(auto &e : arr) {
        cin>>e;
    }
    //ayo wuss poppin
    
    ll ans=0;
    for(int len=3; len<=4; len++) {
        if(len>n) break;
        for(int i=0; i<n-len+1; i++) {
            vector<int> subarr;
            bool ok=1;
            for(int j=i; j<len+i; j++) {
                subarr.pb(arr[j]);
            }
            for(int i=0; i<len; i++) {
                for(int j=i+1; j<len; j++) {
                    for(int k=j+1; k<len; k++) {
                        if(subarr[i]<=subarr[j]&&subarr[j]<=subarr[k]) {
                            ok=0;
                        }
                        if(subarr[i]>=subarr[j]&&subarr[j]>=subarr[k]) {
                            ok=0;
                        }
                    }
                }
            }
            if(ok) {
                ans++;
            }
        }

    }
    ans+=2*n-1;
    cout<<ans<<endl;
}
int main()
{
    fast_cin();
    int t;
    cin>>t;
    while(t--) {
        solve();
    }
}
```