#include <bits/stdc++.h>
using namespace std;
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(num_t lhs, num_t rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t pw(num_t n, num_t k, num_t mod) {
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res;
}
const int inf = 1e9 + 7;
const int mod = inf;
const long long ll_inf = 9ll * inf * inf;
const int MAX_N = 2e5 + 7;
int n;
int a[MAX_N];
int first_index(int u) {
  int cnt = -1;
  while (u > 0) {
    ++cnt;
    u >>= 1;
  }
  return cnt;
}
set<int> matrix[32];
void solve() {
  cin >> n;
  for (int i = (0), _b = (n); i < _b; ++i) {
    cin >> a[i];
    if (i) a[i] ^= a[i - 1];
  }
  if (!a[n - 1]) {
    cout << -1;
    return;
  }
  for (int i = (0), _b = (n); i < _b; ++i)
    if (a[i]) matrix[first_index(a[i])].insert(a[i]);
  int res = 0;
  for (int i = 31; i >= 0; --i)
    if (!matrix[i].empty()) {
      int u = *matrix[i].begin();
      matrix[i].erase(u);
      for (int v : matrix[i]) {
        int cur = u ^ v;
        if (cur) matrix[first_index(cur)].insert(cur);
      }
      matrix[i].clear();
      ++res;
    }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(num_t lhs, num_t rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t pw(num_t n, num_t k, num_t mod) {
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res;
}
const int inf = 1e9 + 7;
const int mod = inf;
const long long ll_inf = 9ll * inf * inf;
const int MAX_N = 2e5 + 7;
int n;
int a[MAX_N];
int first_index(int u) {
  int cnt = -1;
  while (u > 0) {
    ++cnt;
    u >>= 1;
  }
  return cnt;
}
set<int> matrix[32];
void solve() {
  cin >> n;
  for (int i = (0), _b = (n); i < _b; ++i) {
    cin >> a[i];
    if (i) a[i] ^= a[i - 1];
  }
  if (!a[n - 1]) {
    cout << -1;
    return;
  }
  for (int i = (0), _b = (n); i < _b; ++i)
    if (a[i]) matrix[first_index(a[i])].insert(a[i]);
  int res = 0;
  for (int i = 31; i >= 0; --i)
    if (!matrix[i].empty()) {
      int u = *matrix[i].begin();
      matrix[i].erase(u);
      for (int v : matrix[i]) {
        int cur = u ^ v;
        if (cur) matrix[first_index(cur)].insert(cur);
      }
      matrix[i].clear();
      ++res;
    }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

```