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. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<int64_t> a;

void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

vector<vector<int>> pref_cnt_1;

void prepare() {
    pref_cnt_1.assign(61, vector<int>(n + 1, 0));

    vector<int> order(n);
    iota(order.begin(), order.end(), 0);

    for(int bit = 0; bit < 61; bit++) {
        vector<vector<int>> with_bit(2);
        for(int i = 0; i < n; i++) {
            int pos = order[i];
            int cbit = (a[pos] >> bit) & 1;
            with_bit[cbit].push_back(pos);
            pref_cnt_1[bit][i + 1] = pref_cnt_1[bit][i] + cbit;
        }

        order = std::move(with_bit[0]);
        order.insert(order.end(), with_bit[1].begin(), with_bit[1].end());
    }
}

vector<vector<int64_t>> dp;

int64_t rec(int bit, int carry_cnt) {
    if(bit == 61) {
        return carry_cnt == 0;
    }

    int64_t& memo = dp[bit][carry_cnt];
    if(memo != -1) {
        return memo;
    }

    memo = 0;
    for(int x = 0; x < 2; x++) {
        int xor_all = 0, new_carry_cnt = 0;
        for(int v = 0; v < 2; v++) {
            for(int c = 0; c < 2; c++) {
                int all_cnt = c ? carry_cnt : n - carry_cnt;
                int cnt = c ? pref_cnt_1[bit][carry_cnt]
                            : pref_cnt_1[bit][n] - pref_cnt_1[bit][carry_cnt];

                if(v == 0) {
                    cnt = all_cnt - cnt;
                }

                xor_all ^= (x ^ v ^ c) * (cnt & 1);
                if(v - x - c < 0) {
                    new_carry_cnt += cnt;
                }
            }
        }

        if(xor_all == 0) {
            memo += rec(bit + 1, new_carry_cnt);
        }
    }

    return memo;
}

void solve() {
    // Petya wins iff after subtracting x from every pile the Nim XOR of the
    // piles a_i - x is zero. We count, by a bitwise digit DP processing bits
    // from least to most significant, the number of x >= 0 for which all
    // a_i - x XOR to 0.
    //
    // The state is (bit, carry_cnt): carry_cnt is how many piles currently owe
    // a borrow into this bit position from the subtraction so far. The piles
    // are kept in an order where, at each bit, those with an incoming borrow
    // come first; pref_cnt_1[bit] is a prefix count of set bits in that order,
    // so for either borrow group we can read how many piles have bit value 0
    // or 1. For each choice of x's current bit we compute the parity of the XOR
    // of the resulting difference bits across all piles and only keep choices
    // where that parity is 0, accumulating into the next carry_cnt those piles
    // that now generate a borrow (v - x - c < 0).
    //
    // rec counts all x in [0, 2^61) with zero XOR. The valid range is
    // 0 <= x < min(a_i); the only counted x outside this range is x = min,
    // which makes the smallest pile 0 and the rest a_i - min, so we subtract 1
    // when that configuration also XORs to zero.

    prepare();

    dp.assign(61, vector<int64_t>(n + 1, -1));
    int64_t ans = rec(0, 0);

    int64_t mn = *min_element(a.begin(), a.end()), x = 0;
    for(int i = 0; i < n; i++) {
        x ^= a[i] - mn;
    }

    if(x == 0) {
        ans--;
    }

    cout << ans << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution

```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).
