Inversion Count using BIT

#include <bits/stdc++.h>
using namespace std;
const int N =1e5 + 5;
int inv(int pos, int BIT[]){
    int count = 0;
    while(pos > 0){
        count += BIT[pos];
        pos -= (pos & (-pos));
    }
    return count;
}
void upd(int pos, int BIT[], int n){

    while(pos <= n ){
        BIT[pos]++;
        pos += (pos & (-pos));
    }

}
int main(){
       int t;
       scanf("%d", &t);
        while(t--){
            int n, i;long long int ans = 0;
            scanf("%d", &n);
             int BIT[N] = {0};
            vector < int > arr(n), v(n);
            for(i = 0;i < n; i++)
                scanf("%d", &arr[i]), v[i] = arr[i];

            sort(v.begin(), v.end());

            for(i = 0; i < n; i++)
                arr[i] = lower_bound(v.begin(), v.end(), arr[i]) - v.begin() + 1;

            for(i = n-1; i >= 0; i--)
                ans += inv(arr[i]-1, BIT), upd(arr[i], BIT, n);
           cout << ans << '\n';
        }

    return 0;
}