#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long mod = 998244353;
struct basis {
  vector<int> v;
  int d;
  int sz;
  basis() {}
  basis(int _d) {
    d = _d;
    v.assign(d, 0);
    sz = 0;
  }
  basis operator=(const basis& oth) {
    vector<int> z(oth.v);
    v.swap(z);
    d = oth.d;
    sz = oth.sz;
    return *this;
  }
  void init(int _d) {
    d = _d;
    v.assign(d, 0);
    sz = 0;
  }
  void insvec(int mask) {
    for (int i = d - 1; i >= 0; i--) {
      if ((mask & 1 << i) == 0) continue;
      if (!v[i]) {
        v[i] = mask;
        ++sz;
        return;
      }
      mask ^= v[i];
    }
  }
  bool present(int x) {
    for (int i = d - 1; i >= 0; i--) {
      if ((x & 1 << i) == 0) continue;
      if (!v[i]) {
        return false;
      }
      x ^= v[i];
    }
    return true;
  }
  int query(int k) {
    int mask = 0;
    int tot = 1 << sz;
    for (int i = d - 1; i >= 0; i--)
      if (v[i]) {
        int hal = tot / 2;
        if ((k <= hal && (mask & 1 << i) == 0) ||
            (k > hal && (mask & 1 << i) > 0))
          mask ^= v[i];
        if (k > hal) k -= hal;
        tot >>= 1;
      }
    return mask;
  }
};
int n, a;
basis b(30);
void solve() {
  cin >> n;
  int cr = 0;
  while (n--) {
    cin >> a;
    cr = cr ^ a;
    if (cr != 0) b.insvec(cr);
  }
  if (cr == 0)
    cout << -1 << '\n';
  else
    cout << b.sz << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long mod = 998244353;
struct basis {
  vector<int> v;
  int d;
  int sz;
  basis() {}
  basis(int _d) {
    d = _d;
    v.assign(d, 0);
    sz = 0;
  }
  basis operator=(const basis& oth) {
    vector<int> z(oth.v);
    v.swap(z);
    d = oth.d;
    sz = oth.sz;
    return *this;
  }
  void init(int _d) {
    d = _d;
    v.assign(d, 0);
    sz = 0;
  }
  void insvec(int mask) {
    for (int i = d - 1; i >= 0; i--) {
      if ((mask & 1 << i) == 0) continue;
      if (!v[i]) {
        v[i] = mask;
        ++sz;
        return;
      }
      mask ^= v[i];
    }
  }
  bool present(int x) {
    for (int i = d - 1; i >= 0; i--) {
      if ((x & 1 << i) == 0) continue;
      if (!v[i]) {
        return false;
      }
      x ^= v[i];
    }
    return true;
  }
  int query(int k) {
    int mask = 0;
    int tot = 1 << sz;
    for (int i = d - 1; i >= 0; i--)
      if (v[i]) {
        int hal = tot / 2;
        if ((k <= hal && (mask & 1 << i) == 0) ||
            (k > hal && (mask & 1 << i) > 0))
          mask ^= v[i];
        if (k > hal) k -= hal;
        tot >>= 1;
      }
    return mask;
  }
};
int n, a;
basis b(30);
void solve() {
  cin >> n;
  int cr = 0;
  while (n--) {
    cin >> a;
    cr = cr ^ a;
    if (cr != 0) b.insvec(cr);
  }
  if (cr == 0)
    cout << -1 << '\n';
  else
    cout << b.sz << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```