#include<bits/stdc++.h>
#define forn(i,a,b)for(int i=(a),_b=(b);i<=_b;i++)
#define fore(i,b,a)for(int i=(b),_a=(a);i>=_a;i--)
#define rep(i,n)for(int i=0,_n=n;i<n;i++)
#define ll long long
#define pii pair<int,int>
#define m_p make_pair
#define pb push_back
#define fi first
#define se second
#define foreach(i,c) for(__typeof(c.begin())i=(c.begin());i!=(c).end();i++)
#define int ll
using namespace std;
const int N=200010;
int t;
int n;
int a[N];
bool ok(int x,int y,int z){
	int aa=abs(x-y)+abs(a[x]-a[y]);
	int bb=abs(y-z)+abs(a[y]-a[z]);
	int cc=abs(x-z)+abs(a[x]-a[z]);
	if(aa+bb==cc||aa+cc==bb||bb+cc==aa)return 1;
	return 0;
}
signed main(){
	ios::sync_with_stdio(0);
	cin>>t;
	while(t--){
		cin>>n;
		forn(i,1,n)cin>>a[i];
		if(n==1){
			cout<<1<<"\n";
			continue;
		}
		if(n==2){
			cout<<3<<"\n";
			continue;
		}
		ll ans=3;
		forn(i,3,n){
			ll lm=0;
			bool f=0;
			fore(j,i-2,1){
				if(f)break;
				forn(k,j+1,i-1){
					forn(kk,k+1,i)
					if(ok(kk,j,k)){
						lm=j;
						f=1;
						break;
					}
				}
			}
			ans+=(i-lm);
		}
		cout<<ans<<"\n";
	} 
	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<bits/stdc++.h>
#define forn(i,a,b)for(int i=(a),_b=(b);i<=_b;i++)
#define fore(i,b,a)for(int i=(b),_a=(a);i>=_a;i--)
#define rep(i,n)for(int i=0,_n=n;i<n;i++)
#define ll long long
#define pii pair<int,int>
#define m_p make_pair
#define pb push_back
#define fi first
#define se second
#define foreach(i,c) for(__typeof(c.begin())i=(c.begin());i!=(c).end();i++)
#define int ll
using namespace std;
const int N=200010;
int t;
int n;
int a[N];
bool ok(int x,int y,int z){
	int aa=abs(x-y)+abs(a[x]-a[y]);
	int bb=abs(y-z)+abs(a[y]-a[z]);
	int cc=abs(x-z)+abs(a[x]-a[z]);
	if(aa+bb==cc||aa+cc==bb||bb+cc==aa)return 1;
	return 0;
}
signed main(){
	ios::sync_with_stdio(0);
	cin>>t;
	while(t--){
		cin>>n;
		forn(i,1,n)cin>>a[i];
		if(n==1){
			cout<<1<<"\n";
			continue;
		}
		if(n==2){
			cout<<3<<"\n";
			continue;
		}
		ll ans=3;
		forn(i,3,n){
			ll lm=0;
			bool f=0;
			fore(j,i-2,1){
				if(f)break;
				forn(k,j+1,i-1){
					forn(kk,k+1,i)
					if(ok(kk,j,k)){
						lm=j;
						f=1;
						break;
					}
				}
			}
			ans+=(i-lm);
		}
		cout<<ans<<"\n";
	} 
	return 0;
}
```