#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
int t, n, m, k, x, y;
struct Gaussbase2 {
  int numofbits = 31;
  int rk = 0;
  vector<int> Base;
  Gaussbase2() { Base.assign(numofbits, 0); }
  Gaussbase2& operator=(Gaussbase2& g) {
    for (int i = 0; i < numofbits; i++) Base[i] = g.Base[i];
    rk = g.rk;
  }
  bool canbemade(int x) {
    for (int i = numofbits - 1; i >= 0; i--) x = min(x, x ^ Base[i]);
    return x == 0;
  }
  void Add(int x) {
    for (int i = numofbits; i >= 0; i--) {
      if ((x >> i) & 1) {
        if (!Base[i]) {
          Base[i] = x;
          rk++;
          return;
        } else
          x ^= Base[i];
      }
    }
  }
} g;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int tot = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    tot ^= x;
    g.Add(x);
  }
  if (tot == 0)
    cout << -1 << "\n";
  else
    cout << g.rk << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
int t, n, m, k, x, y;
struct Gaussbase2 {
  int numofbits = 31;
  int rk = 0;
  vector<int> Base;
  Gaussbase2() { Base.assign(numofbits, 0); }
  Gaussbase2& operator=(Gaussbase2& g) {
    for (int i = 0; i < numofbits; i++) Base[i] = g.Base[i];
    rk = g.rk;
  }
  bool canbemade(int x) {
    for (int i = numofbits - 1; i >= 0; i--) x = min(x, x ^ Base[i]);
    return x == 0;
  }
  void Add(int x) {
    for (int i = numofbits; i >= 0; i--) {
      if ((x >> i) & 1) {
        if (!Base[i]) {
          Base[i] = x;
          rk++;
          return;
        } else
          x ^= Base[i];
      }
    }
  }
} g;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int tot = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    tot ^= x;
    g.Add(x);
  }
  if (tot == 0)
    cout << -1 << "\n";
  else
    cout << g.rk << "\n";
  return 0;
}

```