#include <bits/stdc++.h>
using namespace std;
namespace LinearBasis {
unsigned long long d[64];
int tot = 0;
void clear() {
  tot = 0;
  memset(d, 0, sizeof(d));
}
bool insert(unsigned long long x) {
  for (int i = 0; i < tot; i++)
    if ((d[i] ^ x) < x) x ^= d[i];
  if (x) {
    d[tot++] = x;
    sort(d, d + tot, [](int x, int y) { return x > y; });
    return 1;
  } else
    return 0;
}
};  // namespace LinearBasis
const int N = 2e5 + 10;
int a[N];
int n;
vector<int> d;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] ^= a[i - 1];
  if (!a[n]) {
    puts("-1");
    return 0;
  }
  int sig = 1;
  d.push_back(0);
  while (sig) {
    sig = 0;
    int l = d[d.size() - 1];
    for (int i = l + 1; i <= n - 1; i++) {
      LinearBasis::clear();
      d.push_back(i);
      d.push_back(n);
      bool tag = 1;
      for (int j = 1; j < d.size() && tag; j++) {
        tag &= LinearBasis::insert(a[d[j]] ^ a[d[j - 1]]);
      }
      if (!tag) {
        d.pop_back();
        d.pop_back();
        continue;
      }
      if (tag) {
        d.pop_back();
        sig = 1;
        break;
      }
    }
  }
  printf("%d\n", d.size());
  return 0;
}

### Prompt

Create a solution in cpp for 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;
namespace LinearBasis {
unsigned long long d[64];
int tot = 0;
void clear() {
  tot = 0;
  memset(d, 0, sizeof(d));
}
bool insert(unsigned long long x) {
  for (int i = 0; i < tot; i++)
    if ((d[i] ^ x) < x) x ^= d[i];
  if (x) {
    d[tot++] = x;
    sort(d, d + tot, [](int x, int y) { return x > y; });
    return 1;
  } else
    return 0;
}
};  // namespace LinearBasis
const int N = 2e5 + 10;
int a[N];
int n;
vector<int> d;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] ^= a[i - 1];
  if (!a[n]) {
    puts("-1");
    return 0;
  }
  int sig = 1;
  d.push_back(0);
  while (sig) {
    sig = 0;
    int l = d[d.size() - 1];
    for (int i = l + 1; i <= n - 1; i++) {
      LinearBasis::clear();
      d.push_back(i);
      d.push_back(n);
      bool tag = 1;
      for (int j = 1; j < d.size() && tag; j++) {
        tag &= LinearBasis::insert(a[d[j]] ^ a[d[j - 1]]);
      }
      if (!tag) {
        d.pop_back();
        d.pop_back();
        continue;
      }
      if (tag) {
        d.pop_back();
        sig = 1;
        break;
      }
    }
  }
  printf("%d\n", d.size());
  return 0;
}

```