#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long LL;
#define FILEIN(s) freopen(s, "r", stdin)
#define FILEOUT(s) freopen(s, "w", stdout)
#define mem(s, v) memset(s, v, sizeof s)

inline int read(void) {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return f * x;
}

const int MAXN = 2e5 + 5;

int n;
int ans;

struct Point {
    int x, y;
}a[MAXN];

inline bool is_bad(const Point&A, const Point&B, const Point&C) {
    int d1 = abs(A.x - B.x) + abs(A.y - B.y);
    int d2 = abs(A.x - C.x) + abs(A.y - C.y);
    int d3 = abs(B.x - C.x) + abs(B.y - C.y);
    return d1 + d2 == d3;
}

inline bool check(const Point&A, const Point&B, const Point&C) {
    if (is_bad(A, B, C)) return false;
    if (is_bad(A, C, B)) return false;
    if (is_bad(B, C, A)) return false;
    return true;
}

int main() {
    int T = read();
    while (T --) {
        n = read();
        ans = n + n - 1;
        for (int i = 1; i <= n; ++ i) {
            a[i].x = read(); a[i].y = i;
        }
        for (int i = 1; i <= n - 2; ++ i) {
            if (check(a[i], a[i + 1], a[i + 2])) ++ ans;
        }
        for (int i = 1; i <= n - 3; ++ i) {
            if (check(a[i], a[i + 1], a[i + 2]) &&
                check(a[i], a[i + 1], a[i + 3]) &&
                check(a[i], a[i + 2], a[i + 3]) &&
                check(a[i + 1], a[i + 2], a[i + 3]))
                    ++ ans;
        }
        printf("%d\n", ans);
    }
    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 <cstdio>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long LL;
#define FILEIN(s) freopen(s, "r", stdin)
#define FILEOUT(s) freopen(s, "w", stdout)
#define mem(s, v) memset(s, v, sizeof s)

inline int read(void) {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return f * x;
}

const int MAXN = 2e5 + 5;

int n;
int ans;

struct Point {
    int x, y;
}a[MAXN];

inline bool is_bad(const Point&A, const Point&B, const Point&C) {
    int d1 = abs(A.x - B.x) + abs(A.y - B.y);
    int d2 = abs(A.x - C.x) + abs(A.y - C.y);
    int d3 = abs(B.x - C.x) + abs(B.y - C.y);
    return d1 + d2 == d3;
}

inline bool check(const Point&A, const Point&B, const Point&C) {
    if (is_bad(A, B, C)) return false;
    if (is_bad(A, C, B)) return false;
    if (is_bad(B, C, A)) return false;
    return true;
}

int main() {
    int T = read();
    while (T --) {
        n = read();
        ans = n + n - 1;
        for (int i = 1; i <= n; ++ i) {
            a[i].x = read(); a[i].y = i;
        }
        for (int i = 1; i <= n - 2; ++ i) {
            if (check(a[i], a[i + 1], a[i + 2])) ++ ans;
        }
        for (int i = 1; i <= n - 3; ++ i) {
            if (check(a[i], a[i + 1], a[i + 2]) &&
                check(a[i], a[i + 1], a[i + 3]) &&
                check(a[i], a[i + 2], a[i + 3]) &&
                check(a[i + 1], a[i + 2], a[i + 3]))
                    ++ ans;
        }
        printf("%d\n", ans);
    }
    return 0;
}
```