#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<long long, long long>;
const double EPS = 1e-9;
const long long INF = 1e18;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
void debug_out(string s, T t) {
  cout << "[" << s << ": " << t << "]\n";
}
template <typename T, typename... U>
void debug_out(string s, T t, U... u) {
  long long w = 0, c = 0;
  while (w < (long long)s.size()) {
    if (s[w] == '(') c++;
    if (s[w] == ')') c--;
    if (!c and s[w] == ',') break;
    w++;
  }
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  debug_out(s.substr(w + 2, (long long)s.size() - w - 1), u...);
}
const long long N = 1e5 + 5, mod = 1e9 + 7;
struct Gauss {
  long long bits, siz = 0;
  vector<long long> basis;
  Gauss(long long bits = 60) : bits(bits), basis(bits) {}
  long long size() { return siz; }
  bool can(long long x) {
    for (long long i = bits - 1; i >= 0; i--) x = min(x, x ^ basis[i]);
    return x == 0;
  }
  void add(long long x) {
    for (long long i = bits - 1; i >= 0; i--) {
      if ((x & 1LL << i) == 0) continue;
      if (!basis[i]) {
        basis[i] = x;
        siz++;
        return;
      }
      x ^= basis[i];
    }
  }
  long long getBest() {
    long long x = 0;
    for (long long i = bits - 1; i >= 0; i--) x = max(x, x ^ basis[i]);
    return x;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, z = 0;
  cin >> n;
  Gauss G;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    z ^= x;
    G.add(x);
  }
  if (!z)
    cout << -1 << "\n";
  else
    cout << G.size() << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

You are given an array a_1, a_2, ..., a_n of integer numbers.

Your task is to divide the array into the maximum number of segments in such a way that:

  * each element is contained in exactly one segment; 
  * each segment contains at least one element; 
  * there doesn't exist a non-empty subset of segments such that bitwise XOR of the numbers from them is equal to 0. 



Print the maximum number of segments the array can be divided into. Print -1 if no suitable division exists.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the array.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).

Output

Print the maximum number of segments the array can be divided into while following the given constraints. Print -1 if no suitable division exists.

Examples

Input


4
5 5 7 2


Output


2


Input


3
1 2 3


Output


-1


Input


3
3 1 10


Output


3

Note

In the first example 2 is the maximum number. If you divide the array into \{[5], [5, 7, 2]\}, the XOR value of the subset of only the second segment is 5 ⊕ 7 ⊕ 2 = 0. \{[5, 5], [7, 2]\} has the value of the subset of only the first segment being 5 ⊕ 5 = 0. However, \{[5, 5, 7], [2]\} will lead to subsets \{[5, 5, 7]\} of XOR 7, \{[2]\} of XOR 2 and \{[5, 5, 7], [2]\} of XOR 5 ⊕ 5 ⊕ 7 ⊕ 2 = 5.

Let's take a look at some division on 3 segments — \{[5], [5, 7], [2]\}. It will produce subsets:

  * \{[5]\}, XOR 5; 
  * \{[5, 7]\}, XOR 2; 
  * \{[5], [5, 7]\}, XOR 7; 
  * \{[2]\}, XOR 2; 
  * \{[5], [2]\}, XOR 7; 
  * \{[5, 7], [2]\}, XOR 0; 
  * \{[5], [5, 7], [2]\}, XOR 5; 



As you can see, subset \{[5, 7], [2]\} has its XOR equal to 0, which is unacceptable. You can check that for other divisions of size 3 or 4, non-empty subset with 0 XOR always exists.

The second example has no suitable divisions.

The third example array can be divided into \{[3], [1], [10]\}. No subset of these segments has its XOR equal to 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<long long, long long>;
const double EPS = 1e-9;
const long long INF = 1e18;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
void debug_out(string s, T t) {
  cout << "[" << s << ": " << t << "]\n";
}
template <typename T, typename... U>
void debug_out(string s, T t, U... u) {
  long long w = 0, c = 0;
  while (w < (long long)s.size()) {
    if (s[w] == '(') c++;
    if (s[w] == ')') c--;
    if (!c and s[w] == ',') break;
    w++;
  }
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  debug_out(s.substr(w + 2, (long long)s.size() - w - 1), u...);
}
const long long N = 1e5 + 5, mod = 1e9 + 7;
struct Gauss {
  long long bits, siz = 0;
  vector<long long> basis;
  Gauss(long long bits = 60) : bits(bits), basis(bits) {}
  long long size() { return siz; }
  bool can(long long x) {
    for (long long i = bits - 1; i >= 0; i--) x = min(x, x ^ basis[i]);
    return x == 0;
  }
  void add(long long x) {
    for (long long i = bits - 1; i >= 0; i--) {
      if ((x & 1LL << i) == 0) continue;
      if (!basis[i]) {
        basis[i] = x;
        siz++;
        return;
      }
      x ^= basis[i];
    }
  }
  long long getBest() {
    long long x = 0;
    for (long long i = bits - 1; i >= 0; i--) x = max(x, x ^ basis[i]);
    return x;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, z = 0;
  cin >> n;
  Gauss G;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    z ^= x;
    G.add(x);
  }
  if (!z)
    cout << -1 << "\n";
  else
    cout << G.size() << "\n";
  return 0;
}

```