#include<bits/stdc++.h>

using namespace std;

#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())

#define x first
#define y second

typedef long long li;
typedef long double ld;
typedef pair<li, li> pt;

const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;

int n;
vector<li> a;

inline bool read() {
	if(!(cin >> n))
		return false;
	a.resize(n);
	fore (i, 0, n)
		cin >> a[i];
	return true;
}

li d(const pt &a, const pt &b) {
	return abs(a.x - b.x) + abs(a.y - b.y);
}

inline void solve() {
	li ans = 0;
	fore (i, 0, n) {
		fore (j, i, n) {
			if (i + 2 <= j) {
				bool ok = true;
				fore (i1, i, j) fore (i2, i1 + 1, j) {
					if (d(pt(a[i1], i1), pt(a[j], j)) == d(pt(a[i1], i1), pt(a[i2], i2)) + d(pt(a[i2], i2), pt(a[j], j)))
						ok = false;
				}
				if (!ok)
					break;
			}
			ans++;
		}
	}
	cout << ans << endl;
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	int tt = clock();
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cout << fixed << setprecision(15);

	int t; cin >> t;
	
	while(t--) {
		read();
		solve();
		
#ifdef _DEBUG
		cerr << "TIME = " << clock() - tt << endl;
		tt = clock();
#endif
	}
	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>

using namespace std;

#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())

#define x first
#define y second

typedef long long li;
typedef long double ld;
typedef pair<li, li> pt;

const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;

int n;
vector<li> a;

inline bool read() {
	if(!(cin >> n))
		return false;
	a.resize(n);
	fore (i, 0, n)
		cin >> a[i];
	return true;
}

li d(const pt &a, const pt &b) {
	return abs(a.x - b.x) + abs(a.y - b.y);
}

inline void solve() {
	li ans = 0;
	fore (i, 0, n) {
		fore (j, i, n) {
			if (i + 2 <= j) {
				bool ok = true;
				fore (i1, i, j) fore (i2, i1 + 1, j) {
					if (d(pt(a[i1], i1), pt(a[j], j)) == d(pt(a[i1], i1), pt(a[i2], i2)) + d(pt(a[i2], i2), pt(a[j], j)))
						ok = false;
				}
				if (!ok)
					break;
			}
			ans++;
		}
	}
	cout << ans << endl;
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	int tt = clock();
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cout << fixed << setprecision(15);

	int t; cin >> t;
	
	while(t--) {
		read();
		solve();
		
#ifdef _DEBUG
		cerr << "TIME = " << clock() - tt << endl;
		tt = clock();
#endif
	}
	return 0;
}
```