#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int arr[N];
int n;
bool can(int x) {
  if (n == 0) return false;
  vector<int> piles(x, 0);
  for (int i = 0; i < n; ++i) {
    piles[i % x] ^= arr[i];
  }
  unordered_set<int> st;
  int sz1 = x / 2;
  int sz2 = x - sz1;
  for (int mask = 1; mask < (1 << sz1); ++mask) {
    int xr = 0;
    for (int i = 0; i < sz1; ++i) {
      if ((mask >> i) & 1) {
        xr ^= piles[i];
      }
    }
    if (xr == 0) return false;
    st.insert(xr);
  }
  for (int mask = 1; mask < (1 << sz2); ++mask) {
    int xr = 0;
    for (int i = 0; i < sz2; ++i) {
      if ((mask >> i) & 1) {
        xr ^= piles[sz1 + i];
      }
    }
    if (xr == 0) return 0;
    if (st.count(xr)) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(time(0));
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 0) {
      --i, --n;
    }
  }
  sort(arr, arr + n);
  for (int i = 32; i >= 1; --i) {
    sort(arr, arr + n);
    if (can(i)) {
      cout << i << '\n';
      return 0;
    }
    for (int j = 0; j < 7; ++j) {
      random_shuffle(arr, arr + n);
      if (can(i)) {
        cout << i << '\n';
        return 0;
      }
    }
  }
  cout << -1;
}

### 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;
const int N = 2e5 + 5;
int arr[N];
int n;
bool can(int x) {
  if (n == 0) return false;
  vector<int> piles(x, 0);
  for (int i = 0; i < n; ++i) {
    piles[i % x] ^= arr[i];
  }
  unordered_set<int> st;
  int sz1 = x / 2;
  int sz2 = x - sz1;
  for (int mask = 1; mask < (1 << sz1); ++mask) {
    int xr = 0;
    for (int i = 0; i < sz1; ++i) {
      if ((mask >> i) & 1) {
        xr ^= piles[i];
      }
    }
    if (xr == 0) return false;
    st.insert(xr);
  }
  for (int mask = 1; mask < (1 << sz2); ++mask) {
    int xr = 0;
    for (int i = 0; i < sz2; ++i) {
      if ((mask >> i) & 1) {
        xr ^= piles[sz1 + i];
      }
    }
    if (xr == 0) return 0;
    if (st.count(xr)) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(time(0));
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 0) {
      --i, --n;
    }
  }
  sort(arr, arr + n);
  for (int i = 32; i >= 1; --i) {
    sort(arr, arr + n);
    if (can(i)) {
      cout << i << '\n';
      return 0;
    }
    for (int j = 0; j < 7; ++j) {
      random_shuffle(arr, arr + n);
      if (can(i)) {
        cout << i << '\n';
        return 0;
      }
    }
  }
  cout << -1;
}

```