#include <bits/stdc++.h>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
template <class T1, class T2>
ostream &operator<<(ostream &out, pair<T1, T2> a) {
  out << "{ " << a.first << ", " << a.second << " }";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> a) {
  out << "[ ";
  for (int i = 0; i < a.size(); ++i) {
    out << a[i];
    if (i + 1 < a.size()) {
      out << ", ";
    }
  }
  out << " ]";
  return out;
}
int msb(int x) {
  int res = -1;
  while (x > 0) {
    x /= 2;
    ++res;
  }
  return res;
}
signed main() {
  doRoutine();
  int n;
  cin >> n;
  vector<int> a(n);
  int xs = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i > 0) {
      a[i] ^= a[i - 1];
    }
  }
  if (a.back() == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (0) cerr << __builtin_popcount(17) << endl;
  if (0) cerr << __builtin_popcount(19) << endl;
  vector<bool> used(n);
  vector<int> res;
  while (true) {
    int i = -1;
    for (int j = 0; j < n; ++j) {
      if (!used[j] && (i == -1 || a[j] > a[i])) {
        i = j;
      }
    }
    if (i == -1 || a[i] == 0) {
      break;
    }
    used[i] = true;
    for (int j = 0; j < n; ++j) {
      if (!used[j] && msb(a[i]) == msb(a[j])) {
        a[j] ^= a[i];
      }
    }
    res.push_back(i);
  }
  if (!used[n - 1]) {
    res.pop_back();
    res.push_back(n - 1);
  }
  cout << res.size() << endl;
  return 0;
}

### Prompt

Generate a CPP solution 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>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
template <class T1, class T2>
ostream &operator<<(ostream &out, pair<T1, T2> a) {
  out << "{ " << a.first << ", " << a.second << " }";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> a) {
  out << "[ ";
  for (int i = 0; i < a.size(); ++i) {
    out << a[i];
    if (i + 1 < a.size()) {
      out << ", ";
    }
  }
  out << " ]";
  return out;
}
int msb(int x) {
  int res = -1;
  while (x > 0) {
    x /= 2;
    ++res;
  }
  return res;
}
signed main() {
  doRoutine();
  int n;
  cin >> n;
  vector<int> a(n);
  int xs = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i > 0) {
      a[i] ^= a[i - 1];
    }
  }
  if (a.back() == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (0) cerr << __builtin_popcount(17) << endl;
  if (0) cerr << __builtin_popcount(19) << endl;
  vector<bool> used(n);
  vector<int> res;
  while (true) {
    int i = -1;
    for (int j = 0; j < n; ++j) {
      if (!used[j] && (i == -1 || a[j] > a[i])) {
        i = j;
      }
    }
    if (i == -1 || a[i] == 0) {
      break;
    }
    used[i] = true;
    for (int j = 0; j < n; ++j) {
      if (!used[j] && msb(a[i]) == msb(a[j])) {
        a[j] ^= a[i];
      }
    }
    res.push_back(i);
  }
  if (!used[n - 1]) {
    res.pop_back();
    res.push_back(n - 1);
  }
  cout << res.size() << endl;
  return 0;
}

```