#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rel register ll
#define rec register char
#define gc getchar
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?-1:*p1++)
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
char buf[1<<23],*p1=buf,*p2=buf;
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) pc('-'),x=-x;
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=2e5;
typedef ll fuko[yuzu|10];
fuko a;
int main() {
  for (int t=read();t--;) {
    int n=read(),i,j,k,l;
    for (i=1;i<=n;++i) a[i]=read();
    ll ans=2*n-1;
    for (i=1;i<=n-2;++i) {
      ans+=!(a[i]<=a[i+1]&&a[i+1]<=a[i+2]||a[i]>=a[i+1]&&a[i+1]>=a[i+2]);
    }
    for (i=1;i<=n-3;++i) {
      int ck=1;
      for (j=i;j<=i+3;++j) {
        for (k=j+1;k<=i+3;++k) {
          for (l=k+1;l<=i+3;++l) {
            if (a[j]<=a[k]&&a[k]<=a[l]||a[j]>=a[k]&&a[k]>=a[l])
              ck=0;
          }
        }
      }
      ans+=ck;
    }
    printf("%lld\n",ans);
  }
}
### 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<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rel register ll
#define rec register char
#define gc getchar
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?-1:*p1++)
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
char buf[1<<23],*p1=buf,*p2=buf;
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) pc('-'),x=-x;
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=2e5;
typedef ll fuko[yuzu|10];
fuko a;
int main() {
  for (int t=read();t--;) {
    int n=read(),i,j,k,l;
    for (i=1;i<=n;++i) a[i]=read();
    ll ans=2*n-1;
    for (i=1;i<=n-2;++i) {
      ans+=!(a[i]<=a[i+1]&&a[i+1]<=a[i+2]||a[i]>=a[i+1]&&a[i+1]>=a[i+2]);
    }
    for (i=1;i<=n-3;++i) {
      int ck=1;
      for (j=i;j<=i+3;++j) {
        for (k=j+1;k<=i+3;++k) {
          for (l=k+1;l<=i+3;++l) {
            if (a[j]<=a[k]&&a[k]<=a[l]||a[j]>=a[k]&&a[k]>=a[l])
              ck=0;
          }
        }
      }
      ans+=ck;
    }
    printf("%lld\n",ans);
  }
}
```