#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1000000000;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(12);
  }
} iosetup;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  return os;
}
struct BinaryBasis {
  vector<ll> basis;
  bool update;
  BinaryBasis() : update(false) {}
  bool add(ll bit) {
    for (auto &p : basis) bit = min(bit, bit ^ p);
    if (bit) {
      basis.push_back(bit);
      return update = true;
    } else
      return false;
  }
  bool check(ll bit) {
    for (auto &p : basis) bit = min(bit, bit ^ p);
    return bit == 0;
  }
  void normalize() {
    if (update) {
      sort(begin(basis), end(basis));
      update = false;
    }
    for (int i = (int)basis.size() - 1; i >= 0; i--) {
      for (int j = i - 1; j >= 0; j--)
        basis[i] = min(basis[i], basis[i] ^ basis[j]);
    }
  }
  bool operator==(BinaryBasis &a) {
    normalize(), a.normalize();
    return basis == a.basis;
  }
  ll get_kth(ll k) {
    if ((1ll << basis.size() <= k)) return -1;
    ll ret = 0;
    if (update) {
      sort(begin(basis), end(basis));
      update = false;
    }
    for (int i = (int)basis.size() - 1; i >= 0; i--) {
      if (k < (1ll << i))
        ret = min(ret, ret ^ basis[i]);
      else {
        k -= 1ll << i;
        ret = max(ret, ret ^ basis[i]);
      }
    }
    return ret;
  }
  size_t size() const { return basis.size(); }
};
signed main() {
  int n;
  cin >> n;
  vector<int> a(n);
  int x = 0;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    x ^= a[i];
  }
  if (x == 0) {
    cout << -1 << endl;
    return 0;
  }
  vector<int> s(n + 1);
  s[0] = 0;
  for (int i = 0; i < (n); i++) s[i + 1] = s[i] ^ a[i];
  vector<int> basis;
  int p = 0;
  for (int i = 0; i < (n); i++) {
    int e = s[i + 1] ^ s[p];
    for (auto &b : basis) chmin(e, e ^ b);
    if (e != 0) {
      p = i + 1;
      basis.push_back(e);
    }
  }
  cout << basis.size() << endl;
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
const int INF = 1000000000;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(12);
  }
} iosetup;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  return os;
}
struct BinaryBasis {
  vector<ll> basis;
  bool update;
  BinaryBasis() : update(false) {}
  bool add(ll bit) {
    for (auto &p : basis) bit = min(bit, bit ^ p);
    if (bit) {
      basis.push_back(bit);
      return update = true;
    } else
      return false;
  }
  bool check(ll bit) {
    for (auto &p : basis) bit = min(bit, bit ^ p);
    return bit == 0;
  }
  void normalize() {
    if (update) {
      sort(begin(basis), end(basis));
      update = false;
    }
    for (int i = (int)basis.size() - 1; i >= 0; i--) {
      for (int j = i - 1; j >= 0; j--)
        basis[i] = min(basis[i], basis[i] ^ basis[j]);
    }
  }
  bool operator==(BinaryBasis &a) {
    normalize(), a.normalize();
    return basis == a.basis;
  }
  ll get_kth(ll k) {
    if ((1ll << basis.size() <= k)) return -1;
    ll ret = 0;
    if (update) {
      sort(begin(basis), end(basis));
      update = false;
    }
    for (int i = (int)basis.size() - 1; i >= 0; i--) {
      if (k < (1ll << i))
        ret = min(ret, ret ^ basis[i]);
      else {
        k -= 1ll << i;
        ret = max(ret, ret ^ basis[i]);
      }
    }
    return ret;
  }
  size_t size() const { return basis.size(); }
};
signed main() {
  int n;
  cin >> n;
  vector<int> a(n);
  int x = 0;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    x ^= a[i];
  }
  if (x == 0) {
    cout << -1 << endl;
    return 0;
  }
  vector<int> s(n + 1);
  s[0] = 0;
  for (int i = 0; i < (n); i++) s[i + 1] = s[i] ^ a[i];
  vector<int> basis;
  int p = 0;
  for (int i = 0; i < (n); i++) {
    int e = s[i + 1] ^ s[p];
    for (auto &b : basis) chmin(e, e ^ b);
    if (e != 0) {
      p = i + 1;
      basis.push_back(e);
    }
  }
  cout << basis.size() << endl;
  return 0;
}

```