#include <bits/stdc++.h>
using namespace std;
#define mxSIZE 200000 + 5
#define pii pair<int, int>
#define S second
#define F first
/*--------------- BINARY INDEXED TREE. //BEGINS -----------*/
vector<int>BIT;
/**
* @brief This is a utility function which calculate
* the sum of array[0 ... idx] using BIT data structure.
*
* @param idx Index tree node.
* @return int
*/
int Sum(int idx){
int _sum = 0;
/* Traversing the ancestors of idx node. */
while(idx > 0){
_sum += BIT[idx]; // Add current element in count.
idx -= (idx & -idx); // Move to 'sum' parent node.
}
// Return count of smaller element on right side.
return _sum;
}
/**
* @brief This function updates a node of Binary Indexed Tree(BIT)
* at given index 'idx'. Given value 'change' is added to BIT[idx] node
* and its ancestors.
*
* @param idx Index tree node.
* @param change Effective difference that needs to be updated.
* @return void
*/
void Update(int idx, int change){
/* Traverse over all ancestors of BIT[idx] and add 'change' */
while(idx < BIT.size()){
BIT[idx] += change; // Add 'change' in current node.
idx += (idx & -idx); // Move to 'update' parent node.
}
}
/*--------------- BINARY INDEXED TREE. //ENDS -----------*/
/**
* @brief This function transforms an array into an array containing
* elements from 1 to n(Preserving the relative order between the elements). For example {-4, 10, 5} --> {1, 3, 2}
*
* @param v Input array.
*/
void transformArray(vector<int> &v){
int sz = v.size();
vector<pii>temp(sz);
for(int i=0; i<sz; i++)
temp[i].F = v[i], temp[i].S = i;
sort(temp.begin(), temp.end());
for(int i=0; i<sz; i++){
v[temp[i].S] = i+1;
}
}
// Main function which gives arrays containing number of smaller elements on the right
// and greater elements on the left.
void solve(vector<int> &v, int n){
// Converting array to an array containing elements from [1, n]
transformArray(v);
BIT.clear();
/**
* Create a Binary Indexed Tree of size equal to
* maxElement + 1. ('+1' so that the elements can directly
* be used as index)
*/
BIT.resize(n+1, 0);
// To store count of smaller elements on right
// of greater elements on left.
vector<int> cntSmallerRight(n), cntGreaterLeft(n);
/* Calculate count of smaller elements on right */
for(int i=n-1; i>=0; i--){
// Get count of elements smaller than v[i].
cntSmallerRight[i] = Sum(v[i]-1);
// Update current element in the BIT
Update(v[i], 1);
}
BIT.clear();
BIT.resize(n+1, 0);
/* Calculate count of greater elements on left */
for(int i=0; i<n; i++){
// Get count of elements greater than v[i].
cntGreaterLeft[i] = i - Sum(v[i]-1);
// Update current element in the BIT
Update(v[i], 1);
}
// cout << "Count of Smaller on right: ";
// for(int i=0; i<n; i++)cout << cntSmallerRight[i] << " ";
// cout<<endl;
// cout << "Count of Greater on left: ";
// for(int i=0; i<n; i++)cout << cntGreaterLeft[i] << " ";
// cout<<"\n\n";
int q;
cin >> q;
while(q--){
int x;
cin >> x;
x--;
int ans = 0;
int right_small = cntSmallerRight[x];
int left_greater = cntGreaterLeft[x];
ans += (right_small * left_greater);
int right_greater = (n - x - 1) - cntSmallerRight[x];
int left_smaller = x - cntGreaterLeft[x];
ans += (right_greater * left_smaller);
cout << ans << endl;
}
}
// Driver Function.
int main(){
int tt = 1;
cin >> tt;
while(tt--){
int n; cin >> n;
vector<int>v(n);
for(int i=0; i<n; i++)cin >> v[i];
solve(v, n);
}
return 0;
}