How Many At Once

Counting inversions by modified merge sort. Both halves are already sorted; tmp collects the merged run.

long long cnt = 0;
int i = lo, j = mid + 1, k = 0;
while (i <= mid && j <= hi) {
    if (a[i] <= a[j]) {
        tmp[k++] = a[i++];
    } else {
        cnt += /* ??? */;
        tmp[k++] = a[j++];
    }
}

Which expression makes cnt the exact number of inversions closed by this merge?