1. Abridged Problem Statement  
Given N (≤100) non-negative integers A₁…Aₙ (each up to 10¹⁸), pick any subsequence (i.e., any subset) of them so that the bitwise XOR of its elements is as large as possible. Output that maximum XOR value.

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2. Detailed Editorial  

We want to maximize XOR over any subset of the input numbers.  XOR forms a vector space over GF(2) on the bit‐level, so the problem reduces to finding the maximum possible linear combination (over GF(2)) of the given “bit‐vectors.”  The standard technique is to build a **basis** of these vectors in decreasing bit‐order via a process akin to Gaussian elimination.  Once we have a basis, we can greedily construct the maximum XOR:

Step A: Build a XOR‐basis (“linear basis”)  
- Maintain an array `base[L]` for bit‐positions L = 0…60, each slot either 0 (empty) or storing one basis vector whose highest set‐bit is L.  
- To **insert** a new number x into the basis:  
  - For bits l from high to low:  
    - If bit l of x is 0, skip.  
    - If bit l of x is 1 and `base[l]` is empty, set `base[l] = x` and stop.  
    - If bit l of x is 1 and `base[l]` is nonzero, replace x with `x XOR base[l]` and continue.  
  - This ensures that at the end, all basis vectors are independent and each occupies a unique leading bit.  

Step B: Compute the maximum XOR you can form  
- Initialize an accumulator `ans = 0`.  
- Iterate l from highest bit to lowest:  
  - If XOR’ing `ans` with `base[l]` increases `ans` (i.e., `(ans XOR base[l]) > ans`), then do `ans ^= base[l]`.  
- The result is the maximum subset‐XOR.

Complexities:  
- Building the basis: O(N · B) where B ≈ 60 (max bits).  
- Querying max XOR: O(B).  
- For N ≤ 100 and B=60 this is instantaneous.

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3. Provided C++ Solution with Line‐by‐Line Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// We will build a xor-basis over bits [0..60].
static const int MAX_BITS = 61;
long long base[MAX_BITS];  // base[l] stores a basis vector with highest bit = l, or 0

// Insert x into the basis
void add_to_basis(long long x) {
    // Try to reduce x by existing basis vectors from high bit down
    for (int l = MAX_BITS - 1; l >= 0; --l) {
        // If x has bit l set
        if (!(x & (1LL << l))) continue;  
        if (base[l] == 0) {
            // No basis vector with leading bit l: we set this one
            base[l] = x;
            return;
        }
        // Otherwise, eliminate bit l from x by xoring with the basis vector
        x ^= base[l];
    }
    // If x becomes 0, it's linearly dependent and not added
}

// After building the basis, find the maximum XOR
long long get_max_xor() {
    long long ans = 0;
    // Greedily try to set bits from high to low
    for (int l = MAX_BITS - 1; l >= 0; --l) {
        if ((ans ^ base[l]) > ans) {
            ans ^= base[l];
        }
    }
    return ans;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;
    // Read all numbers and insert into the basis
    for (int i = 0; i < N; i++) {
        long long x;
        cin >> x;
        add_to_basis(x);
    }
    // Compute and print the maximum subset XOR
    cout << get_max_xor() << "\n";
    return 0;
}
```

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4. Python Solution with Detailed Comments  

```python
import sys

def add_to_basis(x, basis):
    """
    Try to insert integer x into the XOR-basis.
    'basis' is a dict mapping bit-position -> basis vector.
    """
    # Process bits from high to low
    for bit in reversed(range(61)):  # bits 0..60
        if not (x >> bit) & 1:
            continue
        if bit not in basis:
            # Found an empty slot: store x here
            basis[bit] = x
            return
        # Eliminate this bit using existing basis vector
        x ^= basis[bit]
    # If x reduces to 0, it's dependent and we discard it

def get_max_xor(basis):
    """
    Given a basis, greedily build the maximum XOR.
    """
    ans = 0
    # Try to improve ans by XORing with each basis vector in descending bit-order
    for bit in sorted(basis.keys(), reverse=True):
        candidate = ans ^ basis[bit]
        if candidate > ans:
            ans = candidate
    return ans

def main():
    data = sys.stdin.read().split()
    n = int(data[0])
    nums = list(map(int, data[1:]))

    basis = {}
    # Build the basis
    for x in nums:
        add_to_basis(x, basis)

    # Print the maximum subset XOR
    print(get_max_xor(basis))

if __name__ == "__main__":
    main()
```

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5. Compressed Editorial  

- View each number as a bit‐vector of length ≤61.  
- Build an independent set of vectors (“basis”) by iteratively inserting each number: if its highest set bit collides with an existing basis vector, XOR it out and continue; otherwise add it.  
- To get the maximum XOR of any subset, start from 0 and for each basis vector (highest‐bit first), XOR it in if it increases the current value.  
- Time O(N·B + B) with B≈60 suffices for N≤100.