﻿#include<cstdio>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<time.h> 
#include<map>
#include<iomanip>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<string>
#include<cstring>
#include<sstream>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
#define int long long
#define RE register
int lowbit(int x) {
	return x & (-x);
}
int gcd(int a, int b) {
	return b ? gcd(b, a % b) : a;
}
int t, n;
int a[200020];
bool find(int i, int j, int k) {
	if ((abs(i - j) + abs(a[i] - a[j]) + abs(j - k) + abs(a[j] - a[k]) != abs(i - k) + abs(a[i] - a[k]) )
		&& (abs(i - j) + abs(a[i] - a[j]) + abs(i - k) + abs(a[i] - a[k]) != abs(j - k) + abs(a[j] - a[k])) 
		&&( abs(j - k) + abs(a[j] - a[k]) + abs(i - k) + abs(a[i] - a[k]) != abs(i - j) + abs(a[i] - a[j]))) {
		return true;
	}
	else {
		return false;
	}
}
signed main(void){
	cin >> t;
	while (t--) {
		cin >> n;
		for (int i = 1; i <= n; i++) {
			cin >> a[i];
		}
		int ans = 2 * n - 1;
		for (int i = 1; i <= n - 3; i++) {
			int j = i + 1, k = i + 2, p = i + 3;
			if (find(i, j, k) && find(i, j, p) && find(j, k, p) && find(i, k, p)) {
				ans++;
			}
		}
		for (int i = 1; i <= n - 2; i++) {
			int j = i + 1, k = i + 2;
			if (find(i, j, k))	ans++;
		}
		cout << ans << endl;
	}

	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<cstdio>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<time.h> 
#include<map>
#include<iomanip>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<string>
#include<cstring>
#include<sstream>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
#define int long long
#define RE register
int lowbit(int x) {
	return x & (-x);
}
int gcd(int a, int b) {
	return b ? gcd(b, a % b) : a;
}
int t, n;
int a[200020];
bool find(int i, int j, int k) {
	if ((abs(i - j) + abs(a[i] - a[j]) + abs(j - k) + abs(a[j] - a[k]) != abs(i - k) + abs(a[i] - a[k]) )
		&& (abs(i - j) + abs(a[i] - a[j]) + abs(i - k) + abs(a[i] - a[k]) != abs(j - k) + abs(a[j] - a[k])) 
		&&( abs(j - k) + abs(a[j] - a[k]) + abs(i - k) + abs(a[i] - a[k]) != abs(i - j) + abs(a[i] - a[j]))) {
		return true;
	}
	else {
		return false;
	}
}
signed main(void){
	cin >> t;
	while (t--) {
		cin >> n;
		for (int i = 1; i <= n; i++) {
			cin >> a[i];
		}
		int ans = 2 * n - 1;
		for (int i = 1; i <= n - 3; i++) {
			int j = i + 1, k = i + 2, p = i + 3;
			if (find(i, j, k) && find(i, j, p) && find(j, k, p) && find(i, k, p)) {
				ans++;
			}
		}
		for (int i = 1; i <= n - 2; i++) {
			int j = i + 1, k = i + 2;
			if (find(i, j, k))	ans++;
		}
		cout << ans << endl;
	}

	return 0;
}
```