#include <bits/stdc++.h>

using namespace std;
#define rep(i, l, r) for(auto i=(l);i<=(r);++i)
#define per(i, r, l) for(auto i=(r);i>=(l);--i)
#define N 3000300
#define ll long long

int n, m, a[N], b;


inline void solve() {
    cin >> n;
    rep(i, 1, n)cin >> a[i];
    ll ans = n + n - 1;
    auto d = [&](int x1, int y1, int x2, int y2) {
        return abs(x1 - x2) + abs(y1 - y2);
    };
    auto Check = [&](int x1, int y1, int x2, int y2, int x3, int y3) {
        int a = d(x1, y1, x2, y2), b = d(x1, y1, x3, y3), c = d(x2, y2, x3, y3);
        if (a + b == c)return 1;
        if (a + c == b)return 1;
        if (b + c == a)return 1;
        return 0;
    };
    auto check = [&](int l, int r) {
        rep(i, l, r)rep(j, i + 1, r)rep(k, j + 1, r) {
                    if (Check(a[i], i, a[j], j, a[k], k))return 0;
                }
        return 1;
    };
    rep(i, 1, n - 2)if (check(i, i + 2))++ans;
    rep(i, 1, n - 3)if (check(i, i + 3))++ans;
    printf("%lld\n", ans);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--)solve();
}

### 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 <bits/stdc++.h>

using namespace std;
#define rep(i, l, r) for(auto i=(l);i<=(r);++i)
#define per(i, r, l) for(auto i=(r);i>=(l);--i)
#define N 3000300
#define ll long long

int n, m, a[N], b;


inline void solve() {
    cin >> n;
    rep(i, 1, n)cin >> a[i];
    ll ans = n + n - 1;
    auto d = [&](int x1, int y1, int x2, int y2) {
        return abs(x1 - x2) + abs(y1 - y2);
    };
    auto Check = [&](int x1, int y1, int x2, int y2, int x3, int y3) {
        int a = d(x1, y1, x2, y2), b = d(x1, y1, x3, y3), c = d(x2, y2, x3, y3);
        if (a + b == c)return 1;
        if (a + c == b)return 1;
        if (b + c == a)return 1;
        return 0;
    };
    auto check = [&](int l, int r) {
        rep(i, l, r)rep(j, i + 1, r)rep(k, j + 1, r) {
                    if (Check(a[i], i, a[j], j, a[k], k))return 0;
                }
        return 1;
    };
    rep(i, 1, n - 2)if (check(i, i + 2))++ans;
    rep(i, 1, n - 3)if (check(i, i + 3))++ans;
    printf("%lld\n", ans);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--)solve();
}

```