#include <iostream>
using namespace std;

bool notMon(int* arr, int a, int b, int c) {
	return !((arr[a] <= arr[b] && arr[b] <= arr[c]) || (arr[a] >= arr[b] && arr[b] >= arr[c]));
}

int isGood(int* arr, int s, int e) {
	if ((e - s) == 2) {
		if (notMon(arr, s, s+1, e))
			return 1;
		else
			return 0;
	}
	else if ((e - s) == 3) {
		if ( notMon(arr, s, s+1, s+2) && notMon(arr, s,s+1,s+3) && notMon(arr, s,s+2,s+3) && notMon(arr, s+1,s+2,s+3) )
			return 1;
		else 
			return 0;
	}
	else if ((e- s) == 4) {
		if (  notMon(arr, s,s+1,s+2)&& notMon(arr, s,s+1,s+3) && notMon(arr, s,s+1,s+4) && notMon(arr, s,s+2,s+3) && notMon(arr, s,s+2,s+4) 
				&& notMon(arr, s,s+3,s+4) && notMon(arr, s+1,s+2,s+3) && notMon(arr, s+1,s+2,s+4) && notMon(arr, s+1,s+3,s+4) && notMon(arr, s+2,s+3,s+4)    )
					return 1;
		else
			return 0;
	}
	
	return -1;
} 

int main() {
	int nRnd = 0; cin >> nRnd;
	for (int rnd = 0; rnd < nRnd; rnd++) {
		int len; cin >> len;
		int* arr = new int[len];
		for (int i = 0; i < len; i++)
			cin >> arr[i];
		
		int ans = 0;
		ans += len;
		ans += (len - 1);
		for (int k = 2; k < len; k++) {
			ans += isGood(arr, k-2, k);
		}
		for (int k = 3; k < len; k++) {
			ans += isGood(arr, k-3, k);
		}
		for (int k = 4; k < len; k++) {
			ans += isGood(arr, k-4, k);
		}
		cout << ans << endl;
		
		delete[] arr;
	}
	return 0;
}

### Prompt

Please formulate 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 <iostream>
using namespace std;

bool notMon(int* arr, int a, int b, int c) {
	return !((arr[a] <= arr[b] && arr[b] <= arr[c]) || (arr[a] >= arr[b] && arr[b] >= arr[c]));
}

int isGood(int* arr, int s, int e) {
	if ((e - s) == 2) {
		if (notMon(arr, s, s+1, e))
			return 1;
		else
			return 0;
	}
	else if ((e - s) == 3) {
		if ( notMon(arr, s, s+1, s+2) && notMon(arr, s,s+1,s+3) && notMon(arr, s,s+2,s+3) && notMon(arr, s+1,s+2,s+3) )
			return 1;
		else 
			return 0;
	}
	else if ((e- s) == 4) {
		if (  notMon(arr, s,s+1,s+2)&& notMon(arr, s,s+1,s+3) && notMon(arr, s,s+1,s+4) && notMon(arr, s,s+2,s+3) && notMon(arr, s,s+2,s+4) 
				&& notMon(arr, s,s+3,s+4) && notMon(arr, s+1,s+2,s+3) && notMon(arr, s+1,s+2,s+4) && notMon(arr, s+1,s+3,s+4) && notMon(arr, s+2,s+3,s+4)    )
					return 1;
		else
			return 0;
	}
	
	return -1;
} 

int main() {
	int nRnd = 0; cin >> nRnd;
	for (int rnd = 0; rnd < nRnd; rnd++) {
		int len; cin >> len;
		int* arr = new int[len];
		for (int i = 0; i < len; i++)
			cin >> arr[i];
		
		int ans = 0;
		ans += len;
		ans += (len - 1);
		for (int k = 2; k < len; k++) {
			ans += isGood(arr, k-2, k);
		}
		for (int k = 3; k < len; k++) {
			ans += isGood(arr, k-3, k);
		}
		for (int k = 4; k < len; k++) {
			ans += isGood(arr, k-4, k);
		}
		cout << ans << endl;
		
		delete[] arr;
	}
	return 0;
}

```