#include<algorithm>
#include<bitset>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<deque>
#include<iostream>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<vector>
using namespace std;
#define ck(x) cout<<#x<<"="<<x<<'\n';
#define ckf(x,y,z) for(int _=y;_<=z;_++)cout<<#x<<'['<<_<<"]="<<x[_]<<'\n';
#define mm(x,y) memset(x,y,sizeof x);
#define endl '\n'
#define x first
#define y second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
const int inf=0x3f3f3f3f;
const int md=1e9+7;
const int maxn=2e5+5;
const int maxm=1e9;

int n,m;

void init(){}
int f[maxn];
ll a[maxn];
inline void solve(){
    cin>>n;
    int len=1;
    for (int i=1;i<=n;i++)
        cin>>a[i],f[i]=0;
    int ans=0;
    for (int i=2;i<n;i++)
    {
        if ((a[i]-a[i-1])*(a[i]-a[i+1])>0)
        {
            if (a[i]>a[i-1])
                f[i]=1;
            else
                f[i]=-1;
            ans++;
            if (i-2>=1&&f[i-1]==-1&&a[i+1]>a[i-1]&&a[i]>a[i-2])
                ans++;
            if (i-2>=1&&f[i-1]==1&&a[i+1]<a[i-1]&&a[i]<a[i-2])
                ans++;
        }
    }
    ans+=n;
    ans+=n-1;
    cout<<ans<<endl;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("D:/I am so vegetable-cb/a.in", "r", stdin);
    freopen("D:/I am so vegetable-cb/a.out", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int __=1;
    cin>>__;
    // init();
    while(__--){
        solve();
    }
    return 0;
}

### Prompt

In Cpp, your task is to solve 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<algorithm>
#include<bitset>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<deque>
#include<iostream>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<vector>
using namespace std;
#define ck(x) cout<<#x<<"="<<x<<'\n';
#define ckf(x,y,z) for(int _=y;_<=z;_++)cout<<#x<<'['<<_<<"]="<<x[_]<<'\n';
#define mm(x,y) memset(x,y,sizeof x);
#define endl '\n'
#define x first
#define y second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
const int inf=0x3f3f3f3f;
const int md=1e9+7;
const int maxn=2e5+5;
const int maxm=1e9;

int n,m;

void init(){}
int f[maxn];
ll a[maxn];
inline void solve(){
    cin>>n;
    int len=1;
    for (int i=1;i<=n;i++)
        cin>>a[i],f[i]=0;
    int ans=0;
    for (int i=2;i<n;i++)
    {
        if ((a[i]-a[i-1])*(a[i]-a[i+1])>0)
        {
            if (a[i]>a[i-1])
                f[i]=1;
            else
                f[i]=-1;
            ans++;
            if (i-2>=1&&f[i-1]==-1&&a[i+1]>a[i-1]&&a[i]>a[i-2])
                ans++;
            if (i-2>=1&&f[i-1]==1&&a[i+1]<a[i-1]&&a[i]<a[i-2])
                ans++;
        }
    }
    ans+=n;
    ans+=n-1;
    cout<<ans<<endl;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("D:/I am so vegetable-cb/a.in", "r", stdin);
    freopen("D:/I am so vegetable-cb/a.out", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int __=1;
    cin>>__;
    // init();
    while(__--){
        solve();
    }
    return 0;
}

```