#include "stdio.h"
#include "math.h"
#include <algorithm>
#include <vector>
#include <iostream>
#include <map>
#include <string>
#include <set>
#include <cstring>

#define max(a,b) (((a)>(b))?(a):(b))
#define min(a,b) (((a)<(b))?(a):(b))
#define INF 0xFFFFFFFFFF

using namespace std;

long long seq[200010];
long long dp[200010];

int main() {
    int t;
	scanf("%d", &t);
	for (int i = 0; i < t; i++) {
		long long res = 0;
		int n;
		scanf("%d", &n);
		for (int j = 0; j < n; j++) 
			scanf("%lld", &seq[j]);
		dp[0] = 1;
		res += dp[0];
		int cur = 1;
		long long last = seq[0];
		for (int j = 1; j < n; j++) {
            if (j == 1) {
				cur = 2;
				last = seq[1];
				res += 2;
				continue;
			}
		    long long tem = (seq[j] - seq[j - 1]) * (seq[j - 1] - seq[j - 2]);
			if (tem >= 0) {
				res += 2;
				continue;
			}
			tem = 0;
			if (j > 2)
				tem = (seq[j - 2] - seq[j - 3]) * (seq[j - 1] - seq[j - 2]);
			if (j > 2 && seq[j - 1] < seq[j - 2] && seq[j - 3] > seq[j - 1] && tem < 0 && seq[j] < seq[j - 2]) {
				res += 4;
				continue;
			}
			if (j > 2 && seq[j - 1] > seq[j - 2] && seq[j - 3] < seq[j - 1] && tem < 0 && seq[j] > seq[j - 2]) {
				res += 4;
				continue;
			}
			res += 3;
		}
		printf("%lld\n", res);
	}
	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 "stdio.h"
#include "math.h"
#include <algorithm>
#include <vector>
#include <iostream>
#include <map>
#include <string>
#include <set>
#include <cstring>

#define max(a,b) (((a)>(b))?(a):(b))
#define min(a,b) (((a)<(b))?(a):(b))
#define INF 0xFFFFFFFFFF

using namespace std;

long long seq[200010];
long long dp[200010];

int main() {
    int t;
	scanf("%d", &t);
	for (int i = 0; i < t; i++) {
		long long res = 0;
		int n;
		scanf("%d", &n);
		for (int j = 0; j < n; j++) 
			scanf("%lld", &seq[j]);
		dp[0] = 1;
		res += dp[0];
		int cur = 1;
		long long last = seq[0];
		for (int j = 1; j < n; j++) {
            if (j == 1) {
				cur = 2;
				last = seq[1];
				res += 2;
				continue;
			}
		    long long tem = (seq[j] - seq[j - 1]) * (seq[j - 1] - seq[j - 2]);
			if (tem >= 0) {
				res += 2;
				continue;
			}
			tem = 0;
			if (j > 2)
				tem = (seq[j - 2] - seq[j - 3]) * (seq[j - 1] - seq[j - 2]);
			if (j > 2 && seq[j - 1] < seq[j - 2] && seq[j - 3] > seq[j - 1] && tem < 0 && seq[j] < seq[j - 2]) {
				res += 4;
				continue;
			}
			if (j > 2 && seq[j - 1] > seq[j - 2] && seq[j - 3] < seq[j - 1] && tem < 0 && seq[j] > seq[j - 2]) {
				res += 4;
				continue;
			}
			res += 3;
		}
		printf("%lld\n", res);
	}
	return 0;
}
```