#include <bits/stdc++.h>
//
const int INF = 0x3f3f3f3f;//1.06e9大小
const int mod = 1e9+7;
const double PI = acos ( -1 );
const double eps = 1e-6;
typedef unsigned long long ull;
typedef long long ll;
#define pii pair<int ,int>
#define ms(a,k) memset(a,k,sizeof(a))
#define X first
#define Y second
#define lowbit a&(-a)
using namespace std;
const int maxn = 2e5+10;
int n,a[maxn];
bool pend(int l,int r)
{
	for(int i =l;i<=r-2;++i)
	{
		for(int j = i+1;j<=r-1;++j)
		{
			for(int k = j+1;k<=r;++k)
			{
				int b[3] = {abs(a[j]-a[i])+j-i,abs(a[k]-a[j])+k-j,abs(a[k]-a[i])+k-i};
				sort(b,b+3);
				//printf("%d %d %d\n",b[0],b[1],b[2]);
				if(b[2] == b[0]+b[1])return false;
			}
		}
	}
	return true;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		int n;scanf("%d",&n);
		for(int i=1;i<=n;++i)scanf("%d",&a[i]);
		int ans =0;
		for(int i=1;i<=n;++i)
		{//以i出发;
			ans += min(n-i+1,2);
			for(int j = i+2;j<=min(n,i+4);++j)
			{//
				if(pend(i,j))++ans;
				else break;
			}
			//printf("%d\n",ans);
		}
		printf("%d\n",ans);
	}
	return 0;
}

### Prompt

Your task is to create 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>
//
const int INF = 0x3f3f3f3f;//1.06e9大小
const int mod = 1e9+7;
const double PI = acos ( -1 );
const double eps = 1e-6;
typedef unsigned long long ull;
typedef long long ll;
#define pii pair<int ,int>
#define ms(a,k) memset(a,k,sizeof(a))
#define X first
#define Y second
#define lowbit a&(-a)
using namespace std;
const int maxn = 2e5+10;
int n,a[maxn];
bool pend(int l,int r)
{
	for(int i =l;i<=r-2;++i)
	{
		for(int j = i+1;j<=r-1;++j)
		{
			for(int k = j+1;k<=r;++k)
			{
				int b[3] = {abs(a[j]-a[i])+j-i,abs(a[k]-a[j])+k-j,abs(a[k]-a[i])+k-i};
				sort(b,b+3);
				//printf("%d %d %d\n",b[0],b[1],b[2]);
				if(b[2] == b[0]+b[1])return false;
			}
		}
	}
	return true;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		int n;scanf("%d",&n);
		for(int i=1;i<=n;++i)scanf("%d",&a[i]);
		int ans =0;
		for(int i=1;i<=n;++i)
		{//以i出发;
			ans += min(n-i+1,2);
			for(int j = i+2;j<=min(n,i+4);++j)
			{//
				if(pend(i,j))++ans;
				else break;
			}
			//printf("%d\n",ans);
		}
		printf("%d\n",ans);
	}
	return 0;
}

```