INVERSION COUNT USING BIT

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ld long double
#define pb push_back
#define fe first
#define se second
#define st(x) sort(x.begin(),x.end())
#define rst(x) sort(x.rbegin(), x.rend())
#define all(x) x.begin(),x.end()
long double pi = 3.14159265358979323;

const double EPS = 1e-12;
const int N = 2*1e5 + 5;
const int mod = 1e9 + 7;

//range update using BIT
ll bit[N],n,a[N];
void update(ll idx,ll val)
{
    while(idx <= n)
    {
        bit[idx] += val;
        idx += idx & (-idx);
    }
}

ll query(ll idx)
{
    ll sum = 0 ;
    while(idx > 0)
    {
        sum += bit[idx];
        idx -= idx & (-idx);
    }
    return sum;
}
ll convert()
{
    ll temp[n];
    for(int i = 0 ; i < n ;i++ ){
              temp[i] = a[i];
    }
    sort(temp , temp + n);
    for(int i = 0 ; i < n ; i++ )
    {
        a[i] = lower_bound(temp , temp + n ,a[i]) - temp + 1;
    }
    for(int i = 0 ; i <= n ; i++ )
        bit[i] = 0 ;
    ll ans = 0 ;

    for(int i = n - 1 ; i >= 0; i-- )
    {
        ans += query(a[i] - 1);
        update(a[i],1);
    }
    return ans;
}
int main()
{
    fast;
    ll t;
    cin >> t ;

    while(t--)
    {
        cin >> n ;
        for(int i = 0 ; i < n ;i++ ){
            cin >> a[i];
        }

        cout << convert() << "\n";
    }

















    return 0;

}