Catherine received an array of integers as a gift for March 8. Eventually she grew bored with it, and she started calculated various useless characteristics for it. She succeeded to do it for each one she came up with. But when she came up with another one — xor of all pairwise sums of elements in the array, she realized that she couldn't compute it for a very large array, thus she asked for your help. Can you do it? Formally, you need to compute

$$$ (a_1 + a_2) ⊕ (a_1 + a_3) ⊕ … ⊕ (a_1 + a_n) \\\ ⊕ (a_2 + a_3) ⊕ … ⊕ (a_2 + a_n) \\\ … \\\ ⊕ (a_{n-1} + a_n) \\\ $$$

Here x ⊕ y is a bitwise XOR operation (i.e. x ^ y in many modern programming languages). You can read about it in Wikipedia: <https://en.wikipedia.org/wiki/Exclusive_or#Bitwise_operation>.

Input

The first line contains a single integer n (2 ≤ n ≤ 400 000) — the number of integers in the array.

The second line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^7).

Output

Print a single integer — xor of all pairwise sums of integers in the given array.

Examples

Input


2
1 2


Output


3

Input


3
1 2 3


Output


2

Note

In the first sample case there is only one sum 1 + 2 = 3.

In the second sample case there are three sums: 1 + 2 = 3, 1 + 3 = 4, 2 + 3 = 5. In binary they are represented as 011_2 ⊕ 100_2 ⊕ 101_2 = 010_2, thus the answer is 2.

⊕ is the bitwise xor operation. To define x ⊕ y, consider binary representations of integers x and y. We put the i-th bit of the result to be 1 when exactly one of the i-th bits of x and y is 1. Otherwise, the i-th bit of the result is put to be 0. For example, 0101_2   ⊕   0011_2 = 0110_2.
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) { return n | (1LL << i); }
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx4[] = {0, 0, +1, -1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dy8[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline double TriangleAreaWithSide(double a, double b, double c) {
  double s = (a + b + c) / 2;
  double area = sqrt(s * (s - a) * (s - b) * (s - c));
  return area;
}
inline double area3(double x1, double y1, double x2, double y2, double x3,
                    double y3) {
  double A = abs((x1 * y2 + x2 * y3 + x3 * y1) - (y1 * x2 + y2 * x3 + y3 * x1));
  A /= 2.0;
  return A;
}
inline long double degreetoradian(long double x) {
  long double val = PI * x;
  val /= (180.0);
  return val;
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
template <typename T>
inline T gcd(T a, T b) {
  T c;
  while (b) {
    c = b;
    b = a % b;
    a = c;
  }
  return a;
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
inline bool equalTo(double a, double b) {
  if (fabs(a - b) <= eps)
    return true;
  else
    return false;
}
inline bool notEqual(double a, double b) {
  if (fabs(a - b) > eps)
    return true;
  else
    return false;
}
inline bool lessThan(double a, double b) {
  if (a + eps < b)
    return true;
  else
    return false;
}
inline bool lessThanEqual(double a, double b) {
  if (a < b + eps)
    return true;
  else
    return false;
}
inline bool greaterThan(double a, double b) {
  if (a > b + eps)
    return true;
  else
    return false;
}
inline bool greaterThanEqual(double a, double b) {
  if (a + eps > b)
    return true;
  else
    return false;
}
inline string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
struct edge {
  int p, q, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int bit[50];
vector<int> v;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.emplace_back(x);
  }
  long long p = 2, q = 1;
  for (int k = 0; k < 30; k++) {
    vector<int> a;
    for (int j = 0; j < n; j++) {
      a.emplace_back(v[j] % p);
    }
    sort((a).begin(), (a).end());
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      int lo = i + 1, hi = n - 1, x = -1;
      int y = q - a[i];
      int z = p - 1 - a[i];
      while (lo <= hi) {
        int mid = (hi + lo) >> 1;
        if (a[mid] >= y && a[mid] <= z) {
          x = mid;
          hi = mid - 1;
        } else if (a[mid] < y)
          lo = mid + 1;
        else
          hi = mid - 1;
      }
      int tem = -1;
      lo = i + 1, hi = n - 1;
      while (lo <= hi) {
        int mid = (hi + lo) >> 1;
        if (a[mid] >= y && a[mid] <= z) {
          tem = mid;
          lo = mid + 1;
        } else if (a[mid] < y)
          lo = mid + 1;
        else
          hi = mid - 1;
      }
      if (x != -1 && tem != -1) ans += (tem - x + 1);
      x = -1, lo = i + 1, hi = n - 1;
      y = p + q - a[i];
      z = p * 2 - 2 - a[i];
      while (lo <= hi) {
        int mid = (hi + lo) >> 1;
        if (a[mid] >= y && a[mid] <= z) {
          x = mid;
          hi = mid - 1;
        } else if (a[mid] < y)
          lo = mid + 1;
        else
          hi = mid - 1;
      }
      tem = -1, lo = i + 1, hi = n - 1;
      while (lo <= hi) {
        int mid = (hi + lo) >> 1;
        if (a[mid] >= y && a[mid] <= z) {
          tem = mid;
          lo = mid + 1;
        } else if (a[mid] < y)
          lo = mid + 1;
        else
          hi = mid - 1;
      }
      if (x != -1 && tem != -1) ans += (tem - x + 1);
    }
    bit[k] = ans % 2;
    p *= 2;
    q *= 2;
  }
  p = 1;
  long long ans = 0;
  for (int i = 0; i < 30; i++) {
    if (bit[i] == 1) ans += p;
    p *= 2;
  }
  cout << ans << '\n';
}
