#include <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <map>
#include <set>
#include <chrono>
#include <queue>
#include <string>
#include <cassert>
#include <bitset>
#include <stack>
#include <sstream>
#include <utility>
#include <unordered_map>

#define INF 2000000000
#define INFLL 2000000000000000000LL
#define MOD 998244353
#define BASE 77747
#define PI acos(-1.0)
#define MAXLEN 31
 
using namespace std;

string s;
int n;
int t;
int arr[200000];

int check(int a, int b, int c) {
    if (a <= b && b <= c) return false;
    if (a >= b && b >= c) return false;
    return true;
}

int check(int a, int b, int c, int d) {
    return check(a, b, c) && check(a, c, d) && check(a, b, d) && check(b, c, d);
}

int main() {
    scanf("%d", &t);
    while (t--) {
        map <int, int> mp;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
        long long ans = 0;
        ans = n;
        ans += n - 1;
        for (int i = 2; i < n; i++) {
            if (check(arr[i - 2], arr[i - 1], arr[i])) ans++;
        }
        for (int i = 3; i < n; i++) {
            if (check(arr[i - 3], arr[i - 2], arr[i - 1], arr[i])) ans++;
        }
        cout << ans << endl;
    }
    return 0;
}

### Prompt

Your challenge is to write 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 <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <map>
#include <set>
#include <chrono>
#include <queue>
#include <string>
#include <cassert>
#include <bitset>
#include <stack>
#include <sstream>
#include <utility>
#include <unordered_map>

#define INF 2000000000
#define INFLL 2000000000000000000LL
#define MOD 998244353
#define BASE 77747
#define PI acos(-1.0)
#define MAXLEN 31
 
using namespace std;

string s;
int n;
int t;
int arr[200000];

int check(int a, int b, int c) {
    if (a <= b && b <= c) return false;
    if (a >= b && b >= c) return false;
    return true;
}

int check(int a, int b, int c, int d) {
    return check(a, b, c) && check(a, c, d) && check(a, b, d) && check(b, c, d);
}

int main() {
    scanf("%d", &t);
    while (t--) {
        map <int, int> mp;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
        long long ans = 0;
        ans = n;
        ans += n - 1;
        for (int i = 2; i < n; i++) {
            if (check(arr[i - 2], arr[i - 1], arr[i])) ans++;
        }
        for (int i = 3; i < n; i++) {
            if (check(arr[i - 3], arr[i - 2], arr[i - 1], arr[i])) ans++;
        }
        cout << ans << endl;
    }
    return 0;
}

```