1. Abridged Problem Statement  
Given n piles with sizes a₁,…,aₙ. Petya may remove x stones from each pile (0 ≤ x < min(aᵢ)). After removal, the players play normal Nim (first player loses iff the XOR of pile sizes is 0). Count how many x make the resulting XOR = 0.

2. Detailed Editorial  

Overview  
We need the number of x ∈ [0, m–1], where m = min(aᵢ), such that  
   F(x) = a₁⊕a₂⊕…⊕aₙ computed after replacing each aᵢ with (aᵢ–x) equals 0.  

Challenge: Subtraction with a common x introduces borrows that couple bits across all piles. A direct bitwise XOR formula does not apply.

Key observation  
When subtracting x from each aᵢ in binary, each bit position k is affected by:  
  – The k-th bit of x (0 or 1).  
  – The k-th bit of aᵢ.  
  – Whether there is a borrow coming in from bit k–1.  

We can do a digit-DP over bit positions from LSB (bit 0) up to the highest bit (we can stop at ~60, since aᵢ ≤ 10¹⁸).  
State D[k][c] = number of ways to choose the lower k bits of x so that:  
 1. The partial XOR of the k LSBs of all (aᵢ–x) is 0.  
 2. Exactly c piles generate a borrow into bit k.  

Transition:  
For bit k, we choose xₖ∈{0,1}. Each pile i has incoming borrow bᵢ ∈ {0,1}. If bᵢ=1, its effective subtraction at bit k is (aᵢₖ – xₖ –1); otherwise (aᵢₖ – xₖ). From that we determine:  
  a) The parity contribution (for the XOR) at this bit across all piles, which must be 0.  
  b) Which piles produce an outgoing borrow into bit k+1.  

We only need to know how many piles have incoming borrow=0 vs 1, and among each group, how many have aₖ=1 vs 0. Let c = number of piles with incoming borrow=1. Then n–c have borrow=0. Precompute for each bit k a prefix array that, for any reorder where the first c piles are those with borrow=1, tells us how many of those c piles have aₖ=1 (and similarly for the other group). With those counts we can in O(1) compute:  
  – The XOR parity at bit k for chosen xₖ.  
  – The number of outgoing borrows to bit k+1.  

Finally D and transitions over k=0…60 in O(60·n) total time.

Edge case: subtracting x=m may produce a zero pile; the problem forbids x≥m. Also, if after subtracting the minimum value x yields all piles equal, the XOR=0 case x=minimum gives a spurious count we must exclude: if XOR_{i}(aᵢ – m) = 0, we must subtract 1 from the total, because x=m is not allowed. In code we handle that by computing the XOR at x=m separately, and if zero, we decrement the final answer by 1.

Complexities  
 Time O(n·log A), Memory O(n·log A).

3. Provided C++ Solution with Detailed Comments  

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

// Overload printing for pairs and vectors for convenience
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
    return out << p.first << ' ' << p.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
    return in >> p.first >> p.second;
}
template<typename T>
istream& operator>>(istream& in, vector<T>& v) {
    for (auto& x : v) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
    for (auto& x : v) out << x << ' ';
    return out;
}

int n;
vector<int64_t> a;

// Read input: n and array a
void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

// pref_cnt_1[k][i] = among the first i indices in the 'order' array for bit k,
//                 how many have their k-th bit = 1.
vector<vector<int>> pref_cnt_1;

// We build an ordering of piles for each bit k so that we can quickly query
// how many piles among any prefix have bit k = 1.
void prepare() {
    // We plan up to bit 60 (because a[i] ≤ 10^18 < 2^60)
    pref_cnt_1.assign(61, vector<int>(n+1, 0));

    // 'order' will hold a permutation of [0..n-1]
    // At each bit we stable-partition by that bit
    vector<int> order(n);
    iota(order.begin(), order.end(), 0);

    for (int bit = 0; bit < 61; ++bit) {
        // We'll split 'order' into two lists: those with bit=0 and bit=1
        vector<vector<int>> bucket(2);
        bucket[0].reserve(n);
        bucket[1].reserve(n);

        // Build prefix counts for how many 1s in this bit so far
        for (int i = 0; i < n; ++i) {
            int idx = order[i];
            int bitval = (a[idx] >> bit) & 1;
            bucket[bitval].push_back(idx);
            // pref_cnt_1[bit][i+1] = pref_cnt_1[bit][i] + (this pile has bit=1)
            pref_cnt_1[bit][i+1] = pref_cnt_1[bit][i] + bitval;
        }
        // Reassemble 'order': all zero-bit piles first, then one-bit piles
        order = move(bucket[0]);
        order.insert(order.end(), bucket[1].begin(), bucket[1].end());
    }
}

// dp[bit][c] = number of ways to choose bits x_0..x_{bit-1}
// such that:
// 1) The XOR of those lower bits of all (a[i] - x) is 0
// 2) Exactly c piles carry a borrow into bit 'bit'
vector<vector<int64_t>> dp;

// Recursive DP with memoization
int64_t rec(int bit, int carry_cnt) {
    // Base case: we processed bits 0..60
    if (bit == 61) {
        // If no borrows remain, we've formed a valid x
        return (carry_cnt == 0) ? 1 : 0;
    }
    int64_t &memo = dp[bit][carry_cnt];
    if (memo != -1) return memo;
    memo = 0;

    // Try x_bit = 0 or 1
    for (int xbit = 0; xbit < 2; ++xbit) {
        // We must check that the XOR of the resulting bit contributions is 0
        int xor_parity = 0;
        int new_carry = 0;

        // We consider two groups: piles with incoming borrow=0 (size n-carry_cnt)
        // and with borrow=1 (size carry_cnt). For each group, we need to know
        // how many have a[bit]=1 vs 0. Using pref_cnt_1:
        //   Among the first carry_cnt piles in 'order', num_ones = pref_cnt_1[bit][carry_cnt].
        //   Among the other piles, num_ones = pref_cnt_1[bit][n] - pref_cnt_1[bit][carry_cnt].
        int ones_if_borrow1 = pref_cnt_1[bit][carry_cnt];
        int ones_if_borrow0 = pref_cnt_1[bit][n] - ones_if_borrow1;
        int zero_if_borrow1 = carry_cnt - ones_if_borrow1;
        int zero_if_borrow0 = (n - carry_cnt) - ones_if_borrow0;

        // For each combination of (incoming borrow b ∈ {0,1}) and (a_bit v ∈ {0,1}),
        // count how many piles are in that category, then compute:
        //   result_bit = (v - xbit - b) mod 2  → contributes to XOR
        //   borrow_out = 1 if (v - xbit - b) < 0
        // We XOR all result_bit multiplied by (cnt & 1) into xor_parity,
        // and sum all cnt that borrow_out into new_carry.

        // Group (b=0, v=0)
        {
            int cnt = zero_if_borrow0;
            int res = (0 - xbit - 0) & 1;      // bit result
            int br = (0 - xbit < 0) ? cnt : 0; // borrow out
            xor_parity ^= res * (cnt & 1);
            new_carry += br;
        }
        // Group (b=0, v=1)
        {
            int cnt = ones_if_borrow0;
            int res = (1 - xbit) & 1;
            int br = (1 - xbit < 0) ? cnt : 0;
            xor_parity ^= res * (cnt & 1);
            new_carry += br;
        }
        // Group (b=1, v=0)
        {
            int cnt = zero_if_borrow1;
            int res = (0 - xbit - 1) & 1;
            int br = (0 - xbit - 1 < 0) ? cnt : 0;
            xor_parity ^= res * (cnt & 1);
            new_carry += br;
        }
        // Group (b=1, v=1)
        {
            int cnt = ones_if_borrow1;
            int res = (1 - xbit - 1) & 1;
            int br = (1 - xbit - 1 < 0) ? cnt : 0;
            xor_parity ^= res * (cnt & 1);
            new_carry += br;
        }

        // If this bit's XOR is zero, we can recurse
        if (xor_parity == 0) {
            memo += rec(bit+1, new_carry);
        }
    }
    return memo;
}

void solve() {
    // Precompute the ordering and prefix sums
    prepare();

    // Initialize dp to -1 (uncomputed)
    dp.assign(61, vector<int64_t>(n+1, -1));

    // Count all valid x in [0..∞) that satisfy XOR=0
    int64_t answer = rec(0, 0);

    // But we only allow x < min(a[i]). We must check if x = min(a) would
    // also be counted. If so, subtract 1.
    int64_t mn = *min_element(a.begin(), a.end());
    int64_t xor_at_mn = 0;
    for (auto v : a) xor_at_mn ^= (v - mn);
    if (xor_at_mn == 0) --answer;

    cout << answer << "\n";
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
import sys
sys.setrecursionlimit(10**7)

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

def prepare(a, n):
    # pref_cnt_1[k][i] = among the first i piles in 'order' for bit k,
    # how many have bit k = 1
    pref_cnt_1 = [[0]*(n+1) for _ in range(61)]
    order = list(range(n))
    for bit in range(61):
        zero_bucket = []
        one_bucket = []
        cnt1 = 0
        for i, idx in enumerate(order):
            b = (a[idx] >> bit) & 1
            if b:
                one_bucket.append(idx)
            else:
                zero_bucket.append(idx)
            cnt1 += b
            pref_cnt_1[bit][i+1] = cnt1
        # new order: zeros then ones
        order = zero_bucket + one_bucket
    return pref_cnt_1

def solve():
    n, a = read_input()
    pref_cnt_1 = prepare(a, n)

    # dp[bit][carry_cnt]: number of ways from bit..60 with carry_cnt incoming borrows
    from functools import lru_cache
    @lru_cache(None)
    def dp(bit, carry):
        # If past final bit, valid only if no carry remains
        if bit == 61:
            return 1 if carry == 0 else 0
        res = 0
        # Precompute counts for this bit
        total_ones = pref_cnt_1[bit][n]
        ones1 = pref_cnt_1[bit][carry]          # among piles with incoming borrow=1
        zeros1 = carry - ones1
        ones0 = total_ones - ones1             # among piles with incoming borrow=0
        zeros0 = (n - carry) - ones0

        # Try x_bit = 0 or 1
        for xbit in (0,1):
            xor_par = 0
            new_c = 0
            # group b=0,v=0
            cnt = zeros0
            val = (0 - xbit) & 1
            xor_par ^= val * (cnt & 1)
            if 0 - xbit < 0: new_c += cnt
            # group b=0,v=1
            cnt = ones0
            val = (1 - xbit) & 1
            xor_par ^= val * (cnt & 1)
            if 1 - xbit < 0: new_c += cnt
            # group b=1,v=0
            cnt = zeros1
            val = (0 - xbit - 1) & 1
            xor_par ^= val * (cnt & 1)
            if 0 - xbit - 1 < 0: new_c += cnt
            # group b=1,v=1
            cnt = ones1
            val = (1 - xbit - 1) & 1
            xor_par ^= val * (cnt & 1)
            if 1 - xbit - 1 < 0: new_c += cnt

            if xor_par == 0:
                res += dp(bit+1, new_c)
        return res

    # Total ways for x >= 0
    total = dp(0, 0)
    # Subtract invalid x = min(a)
    mn = min(a)
    xor_mn = 0
    for v in a:
        xor_mn ^= (v - mn)
    if xor_mn == 0:
        total -= 1

    print(total)

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

5. Compressed Editorial  

We wish to count x ∈ [0, m–1] so that ⊕ᵢ(aᵢ–x)=0. Subtraction introduces borrows across bits. Use a digit-DP over bit positions 0..60. Define D[k][c] = #ways to choose lower k bits of x achieving zero XOR so far and with c piles carrying a borrow into bit k. For each bit we try xₖ=0/1, compute parity of that bit’s XOR from four categories (borrow in 0/1 × aᵢ’s bit 0/1) and count outgoing borrows. Transition in O(1) with precomputed prefix sums of bit counts over a special ordering. Finally subtract 1 if x=m also led to XOR=0 (x=m is forbidden). Total O(n·60).