Inverse Inversion - 2

Inversion table for a permutation A = (a_1, a_2, \ldots, a_n) of the numbers \{1, 2, \ldots, N\} is defined as an array X = (x_i)_{1 \leq i \leq N}. At the $i$-th position of this array, x_i represents the count of elements greater than $i$ that appear to the left of $i$ in the permutation. In other words, x_i is the number of indices j' such that j' < j and a_{j'} > a_j = i.

For instance, the inversion table for the permutation (2, 5, 1, 3, 4) is (2, 0, 1, 1, 0), and for the permutation (6, 1, 3, 7, 5, 4, 2), it is (1, 5, 1, 3, 2, 0, 0).

The inverse permutation A^{-1} of a permutation $A$ is a permutation where the $i$-th position in A^{-1} indicates the position in $A$ where the element equal to $i$ is located.

For example, for the permutation (2, 5, 1, 3, 4), the inverse permutation is (3, 1, 4, 5, 2) because $1$ is at the third position, $2$ is at the first, $3$ is at the fourth, $4$ is at the fifth, and $5$ is at the second. Similarly, for the permutation (2, 7, 3, 6, 5, 1, 4), the inverse is (6, 1, 3, 7, 5, 4, 2).

Your task is to compute the inversion table of the inverse permutation A^{-1}, given the inversion table of permutation $A$.