<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

542. Gena vs Petya
Time limit per test: 1 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



Gena and Petya love playing the following game with each other. There are n piles of stones, the i-th pile contains ai stones. The players move in turns, Gena moves first. A player moves by choosing any non-empty pile and taking an arbitrary positive number of stones from it. If the move is impossible (that is, all piles are empty), then the game finishes and the current player is considered a loser.

Gena and Petya are the world famous experts in unusual games. We will assume that they play optimally.

Recently Petya started to notice that Gena wins too often. Petya decided that the problem is the unjust rules as Gena always gets to move first! To even their chances, Petya decided to cheat and take and hide some stones before the game begins. Since Petya does not want Gena to suspect anything, he will take the same number of stones x from each pile. This number x can be an arbitrary non-negative integer, strictly less that the minimum of ai values.

Your task is to find the number of distinct numbers x such that Petya will win the game.

Input
The first line contains the number of piles n (1 ≤ n ≤ 2 · 105). The second line contains n space-separated integers ai (1 ≤ ai ≤ 1018) — the piles' sizes.

Output
Print the number of ways to choose x so that Petya will win the resulting game considering that both players play optimally.

Example(s)
sample input
sample output
2
3 3
3

sample input
sample output
3
3 4 5
1

sample input
sample output
4
2 7 4 1
1

sample input
sample output
4
4 6 8 10
2



Note
Consider the first example. Petya can choose any x between 0 and 2. After it Gena starts the game with two piles of equal sizes and looses the game. In the second example there is a single possible value of x, equal to 2. In the third example the sought x is also only one — it's x=0. In the fourth example there are two possible values of x — they are 0 and 3.

<|response|>
1. Abridged problem statement
   - You have n piles of stones with sizes a₁,…,aₙ.
   - Petya may choose an integer x, 0 ≤ x < m, where m = min(aᵢ), and remove x stones from each pile.
   - Then Gena and Petya play standard Nim: they alternate removing any positive number of stones from a single non-empty pile; the player who cannot move loses.
   - In Nim, the first player loses if and only if the bitwise XOR of the pile sizes is 0.
   - Petya moves "before" the game by choosing x so that after subtraction the XOR of (aᵢ–x) is 0—then Gena (the first Nim player) loses.
   - Count the number of valid x ∈ [0, m–1] that make ⊕ᵢ(aᵢ–x)=0.

2. Key observations
   - We need to count x ∈ [0, m–1] with
       XOR(x) ≔ (a₁–x)⊕(a₂–x)⊕…⊕(aₙ–x) = 0.
   - Subtraction of x from each aᵢ in binary involves "borrows" that couple all bits together; you cannot treat each bit independently in a simple XOR formula.
   - We can do a digit-DP on bits 0…60 (since aᵢ ≤ 10¹⁸ < 2⁶⁰).
   - Define DP state:
       D[k][c] = number of ways to choose the k least significant bits of x (bits 0..k–1) so that:
       1. The XOR of bits 0..k–1 of all (aᵢ–x) is 0.
       2. Exactly c piles generate a borrow into bit k.
   - Transition from D[k][c] to D[k+1][*]: try xₖ∈{0,1}. For each pile i, given its k-th bit aᵢₖ and whether it had a borrow in bᵢ∈{0,1}, we compute:
       • The resulting bit contribution rᵢ = (aᵢₖ – xₖ – bᵢ) mod 2.
       • Whether it produces a borrow out of this bit (if aᵢₖ – xₖ – bᵢ < 0).
     We require that the XOR across all piles of rᵢ is 0. The next carry count c′ is the total number of piles with a borrow out.
   - To implement this efficiently, we only need to know, for the current bit k and incoming borrow group sizes c and n–c, how many piles have aᵢₖ=1 in each group. We can precompute, for each bit k, a permutation of piles grouped by their k-th bit and a prefix sum of 1-bits. This lets us in O(1) get the counts needed for all four combinations (borrow-in 0/1 × aᵢₖ 0/1).
   - Finally, the DP will count all x ≥ 0 (with any number of bits). We must subtract 1 if x=m (the forbidden case) also satisfies the XOR=0 condition.

3. Full solution approach
   1. Read n and array a. Compute m = min(a).
   2. Preprocess for each bit k=0..60:
      - Maintain an array order[] initially 0,1,…,n–1.
      - Stable-partition order[] by the k-th bit of a[order[i]] (0-bits first, then 1-bits).
      - While partitioning, build pref1[k][i] = number of 1-bits among the first i piles in this order.
      After this, for any c, the first c indices in order[] represent the piles with borrow-in=1, and pref1[k][c] tells how many of those have aᵢₖ=1; similarly (pref1[k][n]–pref1[k][c]) is how many of the other n–c piles have bit=1.
   3. Run digit-DP with memoization: rec(bit, carry_cnt) returns the number of ways to choose bits from position bit..60 such that the remaining XOR stays zero, starting with carry_cnt piles owing a borrow into bit.
   4. Compute XOR_m = ⊕ᵢ(aᵢ–m). If XOR_m==0, decrement the answer by 1 (to exclude x=m).
   5. Print the result.

4. C++ implementation with comments

```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;
}
```

5. Python implementation with 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()
```
