#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<long long> b(n);
  b[0] = a[0];
  for (int i = 1; i < n; i++) b[i] = b[i - 1] ^ a[i];
  if (b[n - 1] == 0) {
    cout << -1;
    return 0;
  }
  int counter = 1;
  int k = 0;
  while (b[n - 1] >= 1 << k) k++;
  k--;
  for (int i = 0; i < n; i++)
    if (b[i] & (1 << k)) b[i] ^= b[n - 1];
  while (1) {
    long long maxx = 0;
    for (int i = 0; i < n; i++) maxx = max(maxx, b[i]);
    if (maxx == 0) {
      cout << counter;
      return 0;
    }
    counter++;
    int k = 0;
    while (maxx >= 1 << k) k++;
    k--;
    for (int i = 0; i < n; i++)
      if (b[i] & (1 << k)) b[i] ^= maxx;
  }
}

### Prompt

Your challenge is to write 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<long long> b(n);
  b[0] = a[0];
  for (int i = 1; i < n; i++) b[i] = b[i - 1] ^ a[i];
  if (b[n - 1] == 0) {
    cout << -1;
    return 0;
  }
  int counter = 1;
  int k = 0;
  while (b[n - 1] >= 1 << k) k++;
  k--;
  for (int i = 0; i < n; i++)
    if (b[i] & (1 << k)) b[i] ^= b[n - 1];
  while (1) {
    long long maxx = 0;
    for (int i = 0; i < n; i++) maxx = max(maxx, b[i]);
    if (maxx == 0) {
      cout << counter;
      return 0;
    }
    counter++;
    int k = 0;
    while (maxx >= 1 << k) k++;
    k--;
    for (int i = 0; i < n; i++)
      if (b[i] & (1 << k)) b[i] ^= maxx;
  }
}

```