#include <bits/stdc++.h>
using namespace std;
int n, a[200005], ans;
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] ^= a[i - 1];
  }
  if (a[n] == 0) {
    cout << -1;
    return 0;
  }
  int i = 1, j = 0;
  while (i <= n && j < 30) {
    int p = -1;
    for (int k = j; k < 30; k++) {
      if (a[i] & (1 << k)) {
        p = k;
        break;
      }
    }
    if (p == -1) {
      i++;
      continue;
    }
    if (p != j) {
      for (int k = 1; k <= n; k++) {
        a[k] += ((a[k] & (1 << j)) > 0) * ((1 << p) - (1 << j)) +
                (((a[k] & (1 << p)) > 0)) * ((1 << j) - (1 << p));
      }
    }
    for (int k = j + 1; k < 30; k++) {
      if (a[i] & (1 << k)) {
        for (int t = i; t <= n; t++) {
          a[t] ^= (1 << k) * ((a[t] & (1 << j)) > 0);
        }
      }
    }
    ans++;
    i++;
    j++;
  }
  cout << ans;
}

### Prompt

Please provide a Cpp coded solution 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;
int n, a[200005], ans;
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] ^= a[i - 1];
  }
  if (a[n] == 0) {
    cout << -1;
    return 0;
  }
  int i = 1, j = 0;
  while (i <= n && j < 30) {
    int p = -1;
    for (int k = j; k < 30; k++) {
      if (a[i] & (1 << k)) {
        p = k;
        break;
      }
    }
    if (p == -1) {
      i++;
      continue;
    }
    if (p != j) {
      for (int k = 1; k <= n; k++) {
        a[k] += ((a[k] & (1 << j)) > 0) * ((1 << p) - (1 << j)) +
                (((a[k] & (1 << p)) > 0)) * ((1 << j) - (1 << p));
      }
    }
    for (int k = j + 1; k < 30; k++) {
      if (a[i] & (1 << k)) {
        for (int t = i; t <= n; t++) {
          a[t] ^= (1 << k) * ((a[t] & (1 << j)) > 0);
        }
      }
    }
    ans++;
    i++;
    j++;
  }
  cout << ans;
}

```