﻿#include <iostream>
#include <iomanip>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include <deque>
#include <string>
#include <cmath>
#include <bitset>
#include <unordered_set>
#include <unordered_map>

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define range(i, n) for(ll i = 0; i < n; ++i)
#define FOR(i, a, b) for (ll i = a; i <= b; i++)
#define print(a) for (auto& i : a) cout << i << " ";
#define input(a) for (auto& i : a) cin >> i;

//#define _GLIBCXX_DEBUG
//#define DEBUG

using namespace std;

ll d(ll x1, ll y1, ll x2, ll y2) {
	return abs(x1 - x2) + abs(y1 - y2);
}

bool check(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3) {
	return d(x1, y1, x2, y2) + d(x2, y2, x3, y3) == d(x1, y1, x3, y3);
}

int main() {
	ll t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<int> a(n);
		input(a);
		ll ans = 0;
		FOR(i, 0, n - 3) {
			if (!check(0, a[i], 1, a[i + 1], 2, a[i + 2]))
				ans++;
		}
		FOR(i, 0, n - 4) {
			if (!check(0, a[i], 1, a[i + 1], 2, a[i + 2]) && !check(0, a[i], 1, a[i + 1], 3, a[i + 3]) && !check(1, a[i + 1], 2, a[i + 2], 3, a[i + 3]) && !check(0, a[i], 2, a[i + 2], 3, a[i + 3]))
				ans++;
		}
		cout << ans + n + n - 1 << "\n";
	}
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#include <iomanip>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include <deque>
#include <string>
#include <cmath>
#include <bitset>
#include <unordered_set>
#include <unordered_map>

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define range(i, n) for(ll i = 0; i < n; ++i)
#define FOR(i, a, b) for (ll i = a; i <= b; i++)
#define print(a) for (auto& i : a) cout << i << " ";
#define input(a) for (auto& i : a) cin >> i;

//#define _GLIBCXX_DEBUG
//#define DEBUG

using namespace std;

ll d(ll x1, ll y1, ll x2, ll y2) {
	return abs(x1 - x2) + abs(y1 - y2);
}

bool check(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3) {
	return d(x1, y1, x2, y2) + d(x2, y2, x3, y3) == d(x1, y1, x3, y3);
}

int main() {
	ll t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<int> a(n);
		input(a);
		ll ans = 0;
		FOR(i, 0, n - 3) {
			if (!check(0, a[i], 1, a[i + 1], 2, a[i + 2]))
				ans++;
		}
		FOR(i, 0, n - 4) {
			if (!check(0, a[i], 1, a[i + 1], 2, a[i + 2]) && !check(0, a[i], 1, a[i + 1], 3, a[i + 3]) && !check(1, a[i + 1], 2, a[i + 2], 3, a[i + 3]) && !check(0, a[i], 2, a[i + 2], 3, a[i + 3]))
				ans++;
		}
		cout << ans + n + n - 1 << "\n";
	}
	return 0;
}
```